answer: option $(\mathrm{A})$

explanation: because the potential difference between solid sphere and hollow spherical shell depends on the radii of two spheres and charge on inner sphere. here charge and radius both remain constant. therefore, potential difference doesn't change.

so, potential difference between the surface is $V$.

mathematically,

initial potential difference, $\mathrm{V}=\frac{K Q}{r_{a}}-\frac{K Q}{r_{b}}$ where $r_{b}>r_{a}$

Now when the shell is given a charge $-3 Q$ the potential at its surface and also inside will change by $V_{0}=\frac{-3 K Q}{r_{b}}$ so now $V_{\text {sphere }}=K\left[\frac{Q}{r_{a}}-\frac{3 Q}{r_{b}}\right]$

and $V_{\text {shell }}=K\left[\frac{Q}{r_{b}}-\frac{3 Q}{r_{b}}\right]$

so, new potential difference, $\mathrm{V}^{\prime}=K\left[\frac{Q}{r_{\mathrm{a}}}-\frac{Q}{r_{b}}\right]$

here it is clear that initial potential difference = final potential difference