d
As given in the first condition:

Both conducting spheres are shown.

\({V_{in}} - {V_{{\text{out }}}} = \left( {\frac{{{\text{kQ}}}}{{{r_1}}}} \right) - \left( {\frac{{{\text{kQ}}}}{{{{\text{r}}_2}}}} \right)\)

\( = {\text{kQ}}\left( {\frac{1}{{{{\text{r}}_1}}} - \frac{1}{{{{\text{r}}_2}}}} \right) = V\)

In the second condition:

Shell is now given charge \(-4 Q\) 

\({V_{in}} - {V_{out}} = \left( {\frac{{kQ}}{{{r_1}}} - \frac{{4kQ}}{{{r_2}}}} \right) - \left( {\frac{{kQ}}{{{r_2}}} - \frac{{4kQ}}{{{r_2}}}} \right)\)

\( = \frac{{kQ}}{{{r_1}}} - \frac{{kQ}}{{{r_2}}}\)

\(=\mathrm{kQ}\left(\frac{1}{\mathrm{r}_{1}}-\frac{1}{\mathrm{r}_{2}}\right)=\mathrm{V}\)

Hence, we also obtain that potential difference does not depend on charge of outer sphere.

\(\therefore \) \(P.d.\) remains same.