b

$\text{Distance of centre from corner is } \frac{a}{\sqrt 2}$

$\text{Electric Field due to +q is } E_1 = \frac{kq}{\frac{a^2}{2}} = \frac{2kq}{a^2}$

$\text{Resultant Electric Field due to Charges at corner A and C } = 2E_1 = \frac{4kq}{a^2} \text{Towards C.}$

$\text{Resultant Electric Field due to Charges at corner B and B } = 2E_1 = \frac{4kq}{a^2} \text{Towards D}.$

$\text{These two Electric Fields are perpendicular so resultant Electric Field is } = \frac{4kq\sqrt 2}{a^2}$

 $\text{Net Potential at centre is } = 0$