Initially when uncharged shell encloses charge Q, charge distribution due to induction will be as shown,

The potential on surface of inner shell is

$${V_A} = {{kQ} \over a} + {{k( - Q)} \over b} + {{kQ} \over b}$$ ..... (i)

where, k = proportionality constant.

Potential on surface of outer shell is

$${V_B} = {{kQ} \over b} + {{k( - Q)} \over b} + {{kQ} \over b}$$ ..... (ii)

Then, potential difference is

$$\Delta {V_{AB}} = {V_A} - {V_B} = kQ\left( {{1 \over a} - {1 \over b}} \right)$$

Given, $$\Delta {V_{AB}} = V$$

So, $$kQ\left( {{1 \over a} - {1 \over b}} \right) = V$$ ....... (iii)

Finally after giving charge $$-$$ 4Q to outer shell, potential difference will be

$$\Delta {V_{AB}} = {V_A} - {V_B}$$

$$ = \left( {{{kQ} \over a} + {{k( - 4Q)} \over b}} \right) - \left( {{{kQ} \over b} + {{k( - 4Q)} \over b}} \right)$$

$$ = kQ\left( {{1 \over a} - {1 \over b}} \right) = V$$ [from Eq. (iii)]

Hence, we obtain that potential difference does not depend on the charge of outer sphere, hence potential difference remains same.