DThe charge on the capacitor is not conserved.
d
$q=C V \Rightarrow V=q / C$
Due to dielectric insertion, new capacitance
$C_{2}=C K$
Initial energy stored in capacitor, $U_{1}=\frac{q^{2}}{2 C}$
Final energy stored in capacitor, $U_{2}=\frac{q^{2}}{2 K C}$
Change in energy stored, $\Delta U=U_{2}-U_{1}$
$\Delta U=\frac{q^{2}}{2 C}\left(\frac{1}{K}-1\right)=\frac{1}{2} C V^{2}\left(\frac{1}{K}-1\right)$
New potential difference between plates
$V^{\prime}=\frac{q}{C K}=\frac{V}{K}$