The diagram shows the curve $C$ with parametric equations
$x = (1+t)^2, \quad y = (3-t)^2.$ The curve $C$ meets the axes at $(16, 0)$ and $(0, 16)$.

1. Set up an equation for $t$ when $x = 16$ and solve it.

$x = 16$
$(1+t)^2 = 16$
$1+t = 4 \quad \text{or} \quad 1+t = -4$
$t = 3 \quad \text{or} \quad t = -5$

2. Identify which value of $t$ corresponds to point $P$.

When $t = 3$, $y = (3-3)^2 = 0$. This gives the $x$-intercept $(16, 0)$ which is already given in the question.
Thus, the other intersection point $P$ where the line $x = 16$ meets $C$ corresponds to $t = -5$ (shown).

1. Find the $y$-coordinate of the point $P$ where $t = -5$.

At $t = -5$,
$y = (3 - (-5))^2$
$y = 8^2 = 64$
The coordinates of $P$ are $(16, 64)$.

2. Find the gradient of the curve $\frac{\mathrm{d}y}{\mathrm{d}x}$ at $P$ using parametric differentiation.

$\frac{\mathrm{d}x}{\mathrm{d}t} = 2(1+t)$
$\frac{\mathrm{d}y}{\mathrm{d}t} = 2(3-t)(-1) = -2(3-t)$

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}} = \frac{-2(3-t)}{2(1+t)} = -\frac{3-t}{1+t}$

At $t = -5$,
$\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{3-(-5)}{1+(-5)} = -\frac{8}{-4} = 2$

3. Determine the gradient of the normal and formulate its equation.

The gradient of the normal $l$ is $-\frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}} = -\frac{1}{2}$.

Using the point-gradient equation $y - y_1 = m(x - x_1)$:
$y - 64 = -\frac{1}{2}(x - 16)$
$y = -\frac{1}{2}x + 8 + 64$
$y = -\frac{1}{2}x + 72$

1. State or re-derive the Cartesian equation of the normal $l$ (required from part b).

Using the parametric equations $x = (1+t)^2$ and $y = (3-t)^2$:
$\frac{\mathrm{d}x}{\mathrm{d}t} = 2(1+t)$ and $\frac{\mathrm{d}y}{\mathrm{d}t} = -2(3-t)$.
At $P$ where $t = -5$, $\frac{\mathrm{d}x}{\mathrm{d}t} = -8$ and $\frac{\mathrm{d}y}{\mathrm{d}t} = -16 \implies \frac{\mathrm{d}y}{\mathrm{d}x} = 2$.
The gradient of the normal $l$ is $-\frac{1}{2}$.
The coordinates of $P$ are $x = (1-5)^2 = 16$ and $y = (3-(-5))^2 = 64$.
The equation of $l$ is:
$y - 64 = -\frac{1}{2}(x - 16) \implies y = -\frac{1}{2}x + 72$.

2. Find the parameter $t$ at point $Q$ where the line $l$ intersects the curve $C$ again.

Substitute $x = (1+t)^2$ and $y = (3-t)^2$ into $y = -\frac{1}{2}x + 72$:
$(3-t)^2 = -\frac{1}{2}(1+t)^2 + 72$
$9 - 6t + t^2 = -\frac{1}{2}(1 + 2t + t^2) + 72$
Multiply by 2 to eliminate the fraction:
$18 - 12t + 2t^2 = -1 - 2t - t^2 + 144$
$3t^2 - 10t - 125 = 0$
Since the line $l$ meets $C$ at $P(t=-5)$, $(t+5)$ is a known factor:
$(t+5)(3t-25) = 0$
For point $Q$, $t = \frac{25}{3}$.

3. Calculate the boundaries $x = b$ and the corresponding $y$-coordinate for point $Q$.

Substitute $t = \frac{25}{3}$ into the respective parametric equations:
$b = x_Q = \left(1 + \frac{25}{3}\right)^2 = \left(\frac{28}{3}\right)^2 = \frac{784}{9}$
$y_Q = \left(3 - \frac{25}{3}\right)^2 = \left(-\frac{16}{3}\right)^2 = \frac{256}{9}$

4. Calculate the exact area of the region bounded by $l$, $x = 16$, $x = b$, and the $x$-axis.

The region is bounded above by the straight line $l$, forming a trapezium with parallel vertical sides $y_P = 64$ and $y_Q = \frac{256}{9}$. This can be evaluated via definite integration or the geometry area formula:
$\text{Area} = \int_{16}^{\frac{784}{9}} \left( -\frac{1}{2}x + 72 \right) \,\mathrm{d}x = \frac{1}{2} \left( y_P + y_Q \right) \times (x_Q - x_P)$
$\text{Area} = \frac{1}{2} \left( 64 + \frac{256}{9} \right) \left( \frac{784}{9} - 16 \right)$
$= \frac{1}{2} \left( \frac{576 + 256}{9} \right) \left( \frac{784 - 144}{9} \right)$
$= \frac{1}{2} \left( \frac{832}{9} \right) \left( \frac{640}{9} \right)$
$= \frac{416 \times 640}{81} = \frac{266240}{81} \text{ units}^2$

1. Set up the integral for the total bounded area using the geometry of the curves.

The line $l$ meets $C$ at $P(16, 64)$ when $t = -5$, and at $Q(\frac{784}{9}, \frac{256}{9})$ when $t = \frac{25}{3}$.
Between these points, $C$ reaches a minimum $x$-value at $t=-1$, where $x=0$ and $y=16$.
The region is bounded above by the line $l$ (for $x \in [16, \frac{784}{9}]$) and the upper branch of $C$ (for $x \in [0, 16]$), and bounded below by the lower branch of $C$ (for $x \in [0, \frac{784}{9}]$).

Thus, the total area bounded by $C$ and $l$ is:
$\text{Area} = \text{Area under } l + \text{Area under } C_{\text{upper}} - \text{Area under } C_{\text{lower}}$

2. Express the curve areas in terms of the parameter $t$.

From part (c), the area under $l$ is $\frac{266240}{81}$.

For $C_{\text{upper}}$, $t$ ranges from $-5$ to $-1$ as $x$ decreases from $16$ to $0$. Using parametric integration:
$\text{Area under } C_{\text{upper}} = \int_{0}^{16} y_{\text{upper}} \,\mathrm{d}x = \int_{-1}^{-5} y \frac{\mathrm{d}x}{\mathrm{d}t} \,\mathrm{d}t = -\int_{-5}^{-1} y \frac{\mathrm{d}x}{\mathrm{d}t} \,\mathrm{d}t$

For $C_{\text{lower}}$, $t$ ranges from $-1$ to $\frac{25}{3}$ as $x$ increases from $0$ to $\frac{784}{9}$:
$\text{Area under } C_{\text{lower}} = \int_{0}^{784/9} y_{\text{lower}} \,\mathrm{d}x = \int_{-1}^{25/3} y \frac{\mathrm{d}x}{\mathrm{d}t} \,\mathrm{d}t$

3. Combine the integrals to find $c$, $d$ and $\mathrm{f}(t)$.

Substitute the parametric areas into the total area expression:
$\text{Area} = \frac{266240}{81} - \int_{-5}^{-1} y \frac{\mathrm{d}x}{\mathrm{d}t} \,\mathrm{d}t - \int_{-1}^{25/3} y \frac{\mathrm{d}x}{\mathrm{d}t} \,\mathrm{d}t$
$\text{Area} = \frac{266240}{81} - \int_{-5}^{25/3} y \frac{\mathrm{d}x}{\mathrm{d}t} \,\mathrm{d}t = \frac{266240}{81} + \int_{-5}^{25/3} \left(-y \frac{\mathrm{d}x}{\mathrm{d}t}\right) \,\mathrm{d}t$

Comparing this with $\frac{266240}{81} + \int_c^d \mathrm{f}(t) \,\mathrm{d}t$, we find:
$c = -5, \quad d = \frac{25}{3}$

Given $y = (3-t)^2$ and $\frac{\mathrm{d}x}{\mathrm{d}t} = 2(1+t)$:
$\mathrm{f}(t) = -(3-t)^2 \times 2(1+t) = -2(1+t)(3-t)^2$

4. Evaluate the integral to find the required area.

Let $I = \int_{-5}^{25/3} -2(1+t)(3-t)^2 \,\mathrm{d}t$.
Use the substitution $u = 3-t \implies \mathrm{d}t = -\mathrm{d}u$ and $1+t = 4-u$.
When $t = -5$, $u = 8$. When $t = \frac{25}{3}$, $u = -\frac{16}{3}$.

$I = \int_{8}^{-16/3} -2(4-u)u^2 (-\mathrm{d}u) = \int_{8}^{-16/3} (8u^2 - 2u^3) \,\mathrm{d}u$
$I = \left[ \frac{8}{3}u^3 - \frac{1}{2}u^4 \right]_{8}^{-16/3}$

Evaluate at upper limit $\left(-\frac{16}{3}\right)$:
$\frac{8}{3}\left(-\frac{4096}{27}\right) - \frac{1}{2}\left(\frac{65536}{81}\right) = -\frac{32768}{81} - \frac{32768}{81} = -\frac{65536}{81}$

Evaluate at lower limit ($8$):
$\frac{8}{3}(512) - \frac{1}{2}(4096) = \frac{4096}{3} - 2048 = -\frac{2048}{3} = -\frac{55296}{81}$

Subtract to find $I$:
$I = -\frac{65536}{81} - \left(-\frac{55296}{81}\right) = -\frac{10240}{81}$

Finally, the total bounded area is:
$\text{Area} = \frac{266240}{81} - \frac{10240}{81} = \frac{256000}{81}$