$\mathrm{V}_{1}=\frac{\mathrm{q}_{1}}{4 \pi \varepsilon_{0} \mathrm{r}_{1}}$
and due to $\mathrm{B}$ is
$\mathrm{V}_{2}=\frac{\mathrm{q}_{2}}{4 \pi \varepsilon_{0} \mathrm{x}}$
Net potential at $\mathrm{P}$
$\mathrm{V}=\mathrm{V}_{1}+\mathrm{V}_{2}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{\mathrm{q}_{2}}{\mathrm{X}}+\frac{\mathrm{q}_{1}}{\mathrm{r}_{1}}\right)$



