After inserting slab
New capacitance $C'=KC=\large\frac{K \in _0 A}{d}$
New potential difference, $V'=\large\frac{V}{K}$
New electric field $E'= \large\frac{V'}{d}=\frac{V}{kd}$
New charge $Q=C'V'= \large\frac{\in _0 AV}{d}$
Work= Final energy - Initial energy
$\qquad= \large\frac{1}{2}$$ C'V'^2 -\large\frac{1}{2} CV^2$
$\qquad=\large\frac{1}{2} $$ (KC ) \bigg ( 1-\large\frac{1}{K}\bigg)$
=> $ |w| =\large\frac{\in _0 AV^2}{2d} \bigg(1 -\large\frac{1}{k}\bigg)$
Hence C is the correct answer.