An electric dipole is placed as shown in the figure.

The electric potential (in 10² V) at point P due to the dipole is (${\in }_0$ permittivity of free space and $\frac{1}{4{\mathrm{\pi ϵ}}_0}=\mathrm{K}$)

The electric potential at P will be superposition of potential due to +q and –q. (Since the dimensions of dipole are large)
$\Rightarrow {\mathrm{V}}_{\mathrm{p}}={\mathrm{V}}_{-\mathrm{q}}+{\mathrm{V}}_{+\mathrm{q}}$

$=\frac{\mathrm{K}(-\mathrm{q})}{8\times {10}^{-2}}+\frac{\mathrm{K}\left(\mathrm{q}\right)}{2\times {10}^{-2}}$

$=\frac{-100\mathrm{Kq}}{8}+\frac{100\mathrm{Kq}}{2}=\frac{300\mathrm{Kq}}{8}$

$\Rightarrow {\mathrm{V}}_{\mathrm{p}}=\left(\frac{3}{8}\right)\mathrm{qK}\times {10}^2\mathrm{V}$