曲线 $C$ 的参数方程为 $x = 4\sin 2\theta - 2, \quad y = 3 - 4\cos 2\theta \quad \text{for } 0 \le \theta < \pi.$

1. Isolate the trigonometric functions $\sin 2\theta$ and $\cos 2\theta$ from the parametric equations.

$x = 4\sin 2\theta - 2 \implies \sin 2\theta = \frac{x + 2}{4}$

$y = 3 - 4\cos 2\theta \implies \cos 2\theta = \frac{3 - y}{4}$

2. Use the Pythagorean identity to eliminate the parameter $\theta$ and obtain the cartesian equation.

Since $\sin^2 2\theta + \cos^2 2\theta = 1$, we substitute the expressions:

$\left(\frac{x + 2}{4}\right)^2 + \left(\frac{3 - y}{4}\right)^2 = 1$

$\frac{(x + 2)^2}{16} + \frac{(y - 3)^2}{16} = 1$

$(x + 2)^2 + (y - 3)^2 = 16$

3. Determine the geometrical interpretation of the cartesian equation.

The equation $(x + 2)^2 + (y - 3)^2 = 16$ represents a circle.

Check the domain: since $0 \le \theta < \pi$, we have $0 \le 2\theta < 2\pi$, which means the entire circle is traced exactly once.

Centre $= (-2, 3)$
Radius $= \sqrt{16} = 4$

1. Find the coordinates of point $P$ at $\theta = \frac{3}{8}\pi$.

When $\theta = \frac{3}{8}\pi$, $2\theta = \frac{3}{4}\pi$.
$x = 4\sin\left(\frac{3}{4}\pi\right) - 2 = 4\left(\frac{\sqrt{2}}{2}\right) - 2 = 2\sqrt{2} - 2$
$y = 3 - 4\cos\left(\frac{3}{4}\pi\right) = 3 - 4\left(-\frac{\sqrt{2}}{2}\right) = 3 + 2\sqrt{2}$
Coordinates of $P$ are $(2\sqrt{2} - 2, 3 + 2\sqrt{2})$.

2. Find the gradient of the curve at point $P$.

Differentiating parametrically with respect to $\theta$:
$\frac{dx}{d\theta} = 8\cos 2\theta$
$\frac{dy}{d\theta} = 8\sin 2\theta$
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{8\sin 2\theta}{8\cos 2\theta} = \tan 2\theta$

At $2\theta = \frac{3}{4}\pi$, the gradient of the tangent is:
$m_T = \tan\left(\frac{3}{4}\pi\right) = -1$

3. Find the $y$-intercepts of the tangent ($T$) and normal ($N$) to the curve at $P$.

Equation of the tangent at $P$:
$y - (3 + 2\sqrt{2}) = -1\left(x - (2\sqrt{2} - 2)\right)$
To find the intersection with the $y$-axis ($T$), let $x = 0$:
$y = -(-2\sqrt{2} + 2) + 3 + 2\sqrt{2} = 1 + 4\sqrt{2}$
So $T$ is $(0, 1 + 4\sqrt{2})$.

The gradient of the normal is $m_N = -\frac{1}{m_T} = 1$.
Equation of the normal at $P$:
$y - (3 + 2\sqrt{2}) = 1\left(x - (2\sqrt{2} - 2)\right)$
To find intersection with the $y$-axis ($N$), let $x = 0$:
$y = -2\sqrt{2} + 2 + 3 + 2\sqrt{2} = 5$
So $N$ is $(0, 5)$.

4. Calculate the exact area of triangle $NPT$.

Taking the segment $NT$ on the $y$-axis as the base of the triangle:
Length of base $NT = y_T - y_N = (1 + 4\sqrt{2}) - 5 = 4\sqrt{2} - 4$

The corresponding perpendicular height is the $x$-coordinate of $P$ (since $x_P > 0$):
Height $h = x_P = 2\sqrt{2} - 2$

Area of $\triangle NPT = \frac{1}{2} \times \text{base} \times h$
$= \frac{1}{2}(4\sqrt{2} - 4)(2\sqrt{2} - 2)$
$= \frac{1}{2} \times 4(\sqrt{2} - 1) \times 2(\sqrt{2} - 1)$
$= 4(\sqrt{2} - 1)^2$
$= 4(2 - 2\sqrt{2} + 1)$
$= 4(3 - 2\sqrt{2})$
$= 12 - 8\sqrt{2}$