An electric charge 10 ^ - 6 is placed at origin ( 0, 0 ) m of X - Y...

The formula to calculate the potential difference between $P$ and $Q$ is given by

$∆ V = kq\left(\dfrac{1}{{r}_{1}} - \dfrac{1}{{r}_{2}}\right) . . . \left(1\right)$

Now,

$\begin{cases}{r}_{1} & = & \sqrt{( \sqrt{3}{)}^{2} + ( \sqrt{3}{)}^{2}}m \\ & = & \sqrt{6}m\end{cases}$

And,

$\begin{cases}{r}_{2} & = & \sqrt{( \sqrt{6}{)}^{2} + 0}m \\ & = & \sqrt{6}m\end{cases}$

As ${r}_{1} = {r}_{2} = \sqrt{6}m$ and the value of the charge remains constant, the potential difference between the points is zero.