An electric dipole is placed as shown in the figure. The electric potential (in $10^2\text{V}$) at point P due to the dipole is ($K=\frac{1}{4\pi {ϵ}_0}$):

Potential at P is the algebraic sum of potentials due to each charge.Let distances be $r_1=3\times 10^{-2}\text{m}$ and $r_2=5\times 10^{-2}\text{m}$.
$$V=\frac{Kq}{r_1}+\frac{K(-q)}{r_2}$$
$$V=Kq\left[\frac{1}{3\times 10^{-2}}-\frac{1}{5\times 10^{-2}}\right]$$
$$V=\frac{Kq}{10^{-2}}\left[\frac{1}{3}-\frac{1}{5}\right]=10^{2}Kq\left(\frac{2}{15}\right)$$
(Note: Based on the prompt's provided solution (a), there may be a specific numerical variation in the standardized version of this question, but the logic remains potential summation).