Dthe electrostatic energy stored in the capacitor decreases
d
(d)
As plates are moved far away further, capacity of capacitor $\left(C=\frac{\varepsilon_0 A}{d}\right)$ decreases.
As battery remains connected during this activity, potential difference between plates remains same.
So, the charge on plates $(Q=C V)$ decreases.
Also, energy of capacitor $\left(U=\frac{1}{2} C V^2\right)$ decreases as capacitance decreases. So, correct option is $( d )$.