JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Compounds Containing Nitrogen

$KOH$ deprotonates the imide N–H of phthalimide to give the phthalimide anion, which then undergoes $S_{N}2$ alkylation with benzyl chloride to give N-benzylphthalimide (product P).
Counting $\pi$ bonds in N-benzylphthalimide: the benzo ring of the phthalimide contributes 3 C=C $\pi$ bonds, the two carbonyl groups contribute 2 C=O $\pi$ bonds, and the phenyl ring of the newly attached benzyl group contributes 3 more C=C $\pi$ bonds.
Total $=3+2+3=8$ $\pi$ bonds.