1. Given situation

A parallel plate capacitor has dielectric constant $K$ inserted between the plates. Its capacitance in this state is given as $C$, and it is charged to potential $V$.

So initially: $Q = CV$ $U_i = \frac{1}{2}CV^2$

2. When dielectric is removed

If the dielectric-filled capacitance is $C$, then without dielectric the capacitance becomes $C' = \frac{C}{K}$

Since the capacitor is charged first and then the slab is slowly removed, the capacitor is assumed isolated during the process. Hence charge remains constant: $Q = CV$

The new energy after removing dielectric is $U_f = \frac{Q^2}{2C'} = \frac{(CV)^2}{2(C/K)}$ $U_f = \frac{C^2V^2}{2} \cdot \frac{K}{C} = \frac{1}{2}KCV^2$

So work must be done on the capacitor to remove the dielectric: $W_{\text{on}} = U_f - U_i = \frac{1}{2}KCV^2 - \frac{1}{2}CV^2$ $W_{\text{on}} = \frac{1}{2}(K-1)CV^2$

Therefore, work done by the system during removal is $W_{\text{by, remove}} = -\frac{1}{2}(K-1)CV^2$

3. When dielectric is reinserted

Now the dielectric is slowly put back. The capacitor again goes from capacitance $C/K$ to $C$ with the same charge $Q$.

Energy changes from $\frac{1}{2}KCV^2 \to \frac{1}{2}CV^2$

Decrease in energy is $\Delta U = \frac{1}{2}CV^2 - \frac{1}{2}KCV^2 = -\frac{1}{2}(K-1)CV^2$

This means the system does positive work during reinsertion: $W_{\text{by, insert}} = +\frac{1}{2}(K-1)CV^2$

4. Net work done over complete cycle

Total work done by the system: $W_{\text{net}} = W_{\text{by, remove}} + W_{\text{by, insert}}$ $W_{\text{net}} = -\frac{1}{2}(K-1)CV^2 + \frac{1}{2}(K-1)CV^2 = 0$

5. Conclusion

Since the dielectric is removed and then reinserted, the system returns to its initial state. Hence over the complete process, the net work done by the system is: $\boxed{0}$

So the correct option is A.