The correct option is B energy of capacitor will decrease
For parallel plates capacitor, C=Aϵ0d,V=QdAϵ0
Thus capacitance is inversely proportional to separation (d) between the plate and potential is proportional to the separation (d).
After disconnect the battery, the capacitor would maintain its charge indefinitely and if there some leakage current, capacitance will discharge very slowly.
As the energy U=12CV2, so the energy is proportional to separation (d) between the plates. Thus when the plates are brought close , the energy of the capacitor will decrease.