Identifying P. Molecular formula $C_6{H}_6{O}_3$, gives a positive ferric chloride test (so it has a phenolic $-OH$), and has no intramolecular hydrogen bond (so the $-OH$ groups are not adjacent in a way that lets them H-bond intramolecularly). The structure that fits is $1,3,5$-trihydroxybenzene (phloroglucinol), which has three $-OH$ groups meta to each other.

Phloroglucinol $\to$ Q. Phloroglucinol exists in tautomeric equilibrium with its tri-keto form, cyclohexane-$1,3,5$-trione. The triketone reacts with three equivalents of hydroxylamine to give a trioxime $Q$.

Phloroglucinol + excess CH${}_3$I / KOH $\to$ R. With excess methyl iodide and strong base, both $O$-methylation and $C$-methylation occur at the activated $\alpha$-carbons of the triketo tautomer. The fully methylated product $R$ is the $2,2,4,4,6,6$-hexamethylcyclohexane-$1,3,5$-trione (six methyl groups added at the three $\alpha$-positions, two methyls per $\alpha$-carbon).

R + excess iso-butylmagnesium bromide; then H${}_3$O${}^{+}$ $\to$ S. Each of the three carbonyls in $R$ adds one equivalent of the Grignard reagent $(C{H}_3{)}_{2}CHC{H}_{2}MgBr$ followed by aqueous workup, giving three tertiary alcohols on the ring. Each new carbon centre carries one iso-butyl group $-C{H}_{2}CH(C{H}_3{)}_2$ (two methyl groups) and an $-OH$ group.

Counting methyls in S.

From the six $\alpha$-methyl groups already on the ring (from step R): $6$ methyls.

From three iso-butyl groups added by the Grignard, each contributing two methyls: $3\times 2=6$ methyls.

Total: $6+6=12$ methyl groups.