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.