Compound P on oxidation with acidic KMnO _4 gives an acid that is...

For compound P to yield an acid less acidic than benzoic acid upon oxidation, the resulting benzoic acid derivative must contain an electron-donating group (EDG) that destabilizes the carboxylate conjugate base. The $-\text{OCH}_3$ group in $p$-methoxytoluene exerts a $+M$ effect, making $p$-methoxybenzoic acid a weaker acid than benzoic acid. If P were $p$-nitrotoluene, it would yield $p$-nitrobenzoic acid, which is stronger due to the $-M$ effect of $-\text{NO}_2$.

For compound Q to yield an amine less basic than aniline upon reduction, the resulting aniline derivative must contain an electron-withdrawing group (EWG) that decreases electron density on the nitrogen atom. The $-\text{Cl}$ group in $p$-chloronitrobenzene reduces to $p$-chloroaniline, where the $-I$ effect of chlorine makes it a weaker base than aniline. If Q were $p$-nitroanisole or $p$-methylnitrobenzene, they would reduce to $p$-anisidine and $p$-toluidine respectively, both of which are stronger bases than aniline due to the $+M$ and $+I$/hyperconjugation effects of their substituents.

Therefore, P is $p$-methoxytoluene and Q is $p$-chloronitrobenzene.

Answer: P = $p$-Methoxytoluene, Q = $p$-Chloronitrobenzene