The curve C has parametric equations $x = 2t - \dfrac{1}{t^{2}},\quad y = 2t + \dfrac{1}{t},\quad t \in \mathbb{R},\ t \ne 0. The point P on the curve has parameter $t = 1$.

1. Find the coordinates of $P$ by substituting $t = 1$ into the parametric equations.

$x = 2(1) - \frac{1}{1^2} = 2 - 1 = 1$
$y = 2(1) + \frac{1}{1} = 2 + 1 = 3$
The coordinates of $P$ are $(1, 3)$.

2. Differentiate $x$ and $y$ with respect to $t$ and evaluate them at $t = 1$.

$x = 2t - t^{-2} \implies \frac{dx}{dt} = 2 + 2t^{-3}$
At $t = 1$, $\frac{dx}{dt} = 2 + 2(1) = 4$

$y = 2t + t^{-1} \implies \frac{dy}{dt} = 2 - t^{-2}$
At $t = 1$, $\frac{dy}{dt} = 2 - 1 = 1$

3. Calculate the gradient of the tangent, $\frac{dy}{dx}$, at point $P$.

Using the chain rule for parametric equations:
$\frac{dy}{dx} = \frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)} = \frac{1}{4}$

The gradient of the tangent at $P$ is $\frac{1}{4}$.

4. Form the equations of the tangent and the normal.

Equation of the tangent (gradient $m = \frac{1}{4}$):
$y - 3 = \frac{1}{4}(x - 1)$
$4y - 12 = x - 1 \implies 4y = x + 11 \text{ or } y = \frac{1}{4}x + \frac{11}{4}$

Gradient of the normal is $m_n = -\frac{1}{\frac{1}{4}} = -4$.
Equation of the normal:
$y - 3 = -4(x - 1)$
$y - 3 = -4x + 4 \implies y = -4x + 7$

1. Find the coordinates of P and the gradient of the curve at $t=1$ (derived from part i).

$x = 2t - \frac{1}{t^{2}}, \quad y = 2t + \frac{1}{t}$
At $t=1$, $x = 2(1) - 1 = 1$ and $y = 2(1) + 1 = 3$. Thus $P$ is $(1, 3)$.
$\frac{dx}{dt} = 2 + \frac{2}{t^3}, \quad \frac{dy}{dt} = 2 - \frac{1}{t^2}$
At $t=1$, $\frac{dx}{dt} = 4$ and $\frac{dy}{dt} = 1$.
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{1}{4}$

2. Form equations of the tangent and normal to find the coordinates of points A and B (derived from part i).

Equation of the tangent at $P(1,3)$:
$y - 3 = \frac{1}{4}(x - 1) \implies y = \frac{1}{4}x + \frac{11}{4}$
The tangent meets the $y$-axis ($x=0$) at $B$, so $y = \frac{11}{4}$. Thus, $B\left(0, \frac{11}{4}\right)$.

Equation of the normal at $P(1,3)$:
$y - 3 = -4(x - 1) \implies y = -4x + 7$
The normal meets the $x$-axis ($y=0$) at $A$, so $0 = -4x + 7 \implies x = \frac{7}{4}$. Thus, $A\left(\frac{7}{4}, 0\right)$.

3. Calculate the area of quadrilateral OAPB.

The coordinates are $O(0,0)$, $A\left(\frac{7}{4}, 0\right)$, $P(1, 3)$, and $B\left(0, \frac{11}{4}\right)$.
Using the shoelace (array) method to find the area of the polygon:
$\text{Area} = \frac{1}{2} \left| \begin{matrix} 0 & \frac{7}{4} & 1 & 0 & 0 \\ 0 & 0 & 3 & \frac{11}{4} & 0 \end{matrix} \right|$
$\text{Area} = \frac{1}{2} \left| \left( 0 \cdot 0 + \frac{7}{4} \cdot 3 + 1 \cdot \frac{11}{4} + 0 \cdot 0 \right) - \left( 0 \cdot \frac{7}{4} + 0 \cdot 1 + 3 \cdot 0 + \frac{11}{4} \cdot 0 \right) \right|$
$\text{Area} = \frac{1}{2} \left| \frac{21}{4} + \frac{11}{4} \right|$
$\text{Area} = \frac{1}{2}(8) = 4$