We can first draw the quadrilateral first.
$\therefore$ The area of quadrilateral $\text{PQRS} = 2 \times \text{area of } \triangle \text{PQS}$
$\qquad \qquad \qquad = 2 \times \frac{1}{2} \times \text{Base} \times \text{Height}$
$\qquad \qquad \qquad = \text{PQ} \times \text{TS} = 4 \times 2 = 8\;\text{unit}^{2}.$
Correct Answer $:\text{C}$
$\textbf{PS:}$ Consider the $xy$-plane, and suppose $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$ are two points in it. Then the distance between $P_1$ and $P_2$ is $${\color{Green}{d(P_1, P_2) = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.}}$$