Identify each major product and count its sp${}^2$/sp carbons.

P (Hell-Volhard-Zelinsky on PhCH${}_2$COOH): $\alpha$-bromination gives Ph-CH${}_2$-CHBr-COOH. Unsaturated carbons = 6 (the benzene ring).

Q (crossed aldol of PhCHO with CH${}_3$CHO at 293 K): dehydration gives cinnamaldehyde, Ph-CH=CH-CHO. Unsaturated carbons = 6 (ring) + 2 (C=C) + 1 (C=O) = 9.

R (phenylacetylene + allyl bromide via NaNH${}_2$, then Hg${}^{2+}$/H${}_3$O${}^{+}$): alkylation gives Ph-C$\equiv$C-CH${}_2$-CH=CH${}_2$; Markovnikov hydration places the carbonyl next to the phenyl: Ph-CO-CH${}_2$-CH${}_2$-CH=CH${}_2$. Unsaturated carbons = 6 (ring) + 1 (C=O) + 2 (C=C) = 9.

S (2-methylindene reaction sequence): ozonolysis of the cyclopentene C=C in the indene ring breaks it to give an ortho-disubstituted benzene bearing CHO and COCH${}_3$ side chains (a 1,2-disubstituted benzene-acetaldehyde and methyl ketone). Two equivalents of CH${}_3$MgBr add to both carbonyls; subsequent H${}^{+}$/heat dehydrates both alcohols to alkenes. The final product is an ortho-disubstituted benzene with two alkene side chains (one vinyl and one 2-methylprop-1-enyl, i.e. CH=CH${}_2$ and CH=C(CH${}_3$)${}_2$ or analogous). Unsaturated carbons = 6 (ring) + 2 (C=C in first side chain) + 2 (C=C in second side chain) = 10.

$\therefore$ S has the most unsaturated carbons. Option (D).