Capacitance of both capacitors are $C_a=\frac{K{\epsilon }_{0}A}{d}$ and $C_b=\frac{{\epsilon }_{0}A}{d+\left(\frac{d}{K}\right)}=\frac{{\epsilon }_{0}AK}{d\left(K+1\right)}$

Now, charge stored is $q_{\mathrm{a}}=\frac{K{\epsilon }_{0}A}{d}V$ and $q_b=\frac{{\epsilon }_{0}AKV}{d\left(K+1\right)}$.

Electric field between each parallel plate capacitor is $E_a=\frac{q_a}{KA{\epsilon }_0}=\frac{K{\epsilon }_{0}AV}{dK{\epsilon }_{0}A}$$=\frac{V}{d}$ and

${\left(E_b\right)}_{\mathrm{dielectric}}=\frac{E_{\mathrm{air}}}{K}$$=\frac{q_b}{KA{\epsilon }_0}$$=\frac{{\epsilon }_{0}AKV}{d\left(K+1\right)\left(KA{\epsilon }_0\right)}$$=\frac{V}{d\left(K+1\right)}$

So, from above relations, we can see that capacitance decrease by a factor of $\frac{1}{K+1}$.

Work done in the process $W=U_f-U_i$

$=\frac{1}{2}\left(C_f-C_i\right)V^2$

$=\frac{1}{2}\left(\frac{{\epsilon }_{0}AK}{d\left(K+1\right)}-\frac{K{\epsilon }_{0}A}{d}\right)V^2$

$=\frac{1}{2}{V}^2\frac{{\epsilon }_{0}AK}{d}\left(\frac{1}{K+1}-1\right)$

$=\frac{1}{2}\frac{{\epsilon }_{0}AK{V}^2}{d}\frac{1-K-1}{K+1}$

$=\frac{1}{2}\frac{{\epsilon }_{0}A{V}^2}{d}\left(\frac{-K^2}{K+1}\right)$

Thus, the work done in the process depends upon the dielectric material.