I am suspicious of the answer given above by Sam.

Suppose the radius of the shell is $a$. Then, the contribution of the shell, $V_{shell}=\frac{kx}{r}$, to the total potential is only true for $r>a$, i.e., **outside** the shell. Now, the problem says that the shell is grounded, and that means that $V=0$ **inside** the shell. So we cannot simply set $V=\frac{kx}{r}+\frac{kq}{r}$ to zero.