The diagram shows a trapezium with vertices $A(-3, 10)$, $B(0, p)$, $C(1, 6)$ and $D$. The sides $AB$ and $DC$ are parallel and the angle $BAD$ is a right angle. Angle $ABE$ is equal to angle $CBE$.

1. Express the gradient of line $AB$ in terms of $p$.

Gradient of $AB = \frac{p - 10}{0 - (-3)} = \frac{p - 10}{3}$

2. Express the gradient of line $CB$ in terms of $p$.

Gradient of $CB = \frac{6 - p}{1 - 0} = 6 - p$

3. Use the angle property to relate the gradients and solve for $p$.

Since $B$ and $E$ both lie on the $y$-axis, the line $BE$ is the $y$-axis.
Given $\angle ABE = \angle CBE$, the lines $AB$ and $CB$ make equal angles with the vertical $y$-axis.
Therefore, their gradients must be equal in magnitude but opposite in sign (reflections across the $y$-axis).
$m_{AB} = -m_{CB}$
$\frac{p - 10}{3} = -(6 - p)$
$\frac{p - 10}{3} = p - 6$
$p - 10 = 3p - 18$
$2p = 8$
$p = 4$

1. Find the equation of line $DC$ and the coordinates of $D$ (from part b).

Gradient of $AB = \frac{10 - 4}{-3 - 0} = -2$.
Since $AB$ is parallel to $DC$, the gradient of $DC$ is $-2$.
Equation of $DC$ passing through $C(1, 6)$:
$y - 6 = -2(x - 1) \implies y = -2x + 8$
To find $D$, solve the equations of $AD$ and $DC$ simultaneously:
$\frac{1}{2}x + \frac{23}{2} = -2x + 8$
$\frac{5}{2}x = -\frac{7}{2} \implies x = -1.4$
$y = -2(-1.4) + 8 = 10.8$
So, $D$ is $(-1.4, 10.8)$.

2. Apply the shoelace formula to find the area of trapezium $ABCD$.

Using the vertices in anti-clockwise order: $A(-3, 10)$, $D(-1.4, 10.8)$, $C(1, 6)$, $B(0, 4)$:
$\text{Area} = \frac{1}{2} \left| \begin{matrix} -3 & -1.4 & 1 & 0 & -3 \\ 10 & 10.8 & 6 & 4 & 10 \end{matrix} \right|$
$\text{Area} = \frac{1}{2} | (-3)(10.8) + (-1.4)(6) + (1)(4) + (0)(10) - (10(-1.4) + 10.8(1) + 6(0) + 4(-3)) |$
$\text{Area} = \frac{1}{2} | -32.4 - 8.4 + 4 + 0 - (-14 + 10.8 + 0 - 12) |$
$\text{Area} = \frac{1}{2} | -36.8 - (-15.2) |$
$\text{Area} = \frac{1}{2} | -21.6 | = 10.8$