Answer
The electric potential at point $P$ due to the charged particles is $~~3.29\times 10^{-4}~V$
Work Step by Step
We can write a general expression for the electric potential at a point due to a system of charged particles:
$V = \frac{1}{4\pi~\epsilon_0}~\sum \frac{q_i}{r_i}$
By symmetry, the electric potential due to $-q_3$ and $+q_3$ cancel out.
We can find the distance from the corner of the rectangle to the point $P$:
$\sqrt{d^2+(\frac{d}{2})^2} = 1.12~d$
We can find the electric potential at point $P$ due to the charged particles:
$V = \frac{1}{4\pi~\epsilon_0}~\sum \frac{q_i}{r_i}$
$V = \frac{1}{4\pi~\epsilon_0}~(\frac{2q_1}{1.12~d}-\frac{2q_2}{d/2})$
$V = \frac{1}{(4\pi)~(8.854\times 10^{-12}~C/V~m)}~[\frac{(2)(5.00\times 10^{-15}~C)}{(1.12)(0.0254~m)}-\frac{(4)(2.00\times 10^{-15}~C)}{0.0254~m}]$
$V = 3.29\times 10^{-4}~V$
The electric potential at point $P$ due to the charged particles is $~~3.29\times 10^{-4}~V$.
