Step 1: Formation of P. Hydroboration-oxidation of styrene (${\text{PhCH=CH}}_2$) is anti-Markovnikov, giving the primary alcohol ${\text{PhCH}}_2{\text{CH}}_2\text{OH}$ as P (2-phenylethan-1-ol).

Step 2: Pathway P $\to$ Q. Oxidation with CrO${}_3$/H${}_2$SO${}_4$ converts the primary alcohol to phenylacetic acid, ${\text{PhCH}}_2\text{COOH}$. Subsequent treatment with Cl${}_2$ and red phosphorus (Hell-Volhard-Zelinsky reaction) introduces a chlorine at the $\alpha$-carbon: $${\text{PhCH}}_2\text{COOH}\stackrel{{\text{Cl}}_2/\text{Red P}}{\to }\text{PhCH(Cl)COOH}.$$ Thus Q = PhCH(Cl)COOH.

Step 3: Pathway P $\to$ R. SOCl${}_2$ converts the alcohol to ${\text{PhCH}}_2{\text{CH}}_2\text{Cl}$. NaCN displaces chloride to give ${\text{PhCH}}_2{\text{CH}}_2\text{CN}$. Acidic hydrolysis converts the nitrile to the carboxylic acid: $${\text{PhCH}}_2{\text{CH}}_2\text{CN}\stackrel{{\text{H}}_3{\text{O}}^{+},\Delta }{\to }{\text{PhCH}}_2{\text{CH}}_2\text{COOH}.$$ Thus R = 3-phenylpropionic acid.

Step 4: R $\to$ S. Treatment with conc. H${}_2$SO${}_4$ promotes intramolecular Friedel-Crafts acylation: the protonated carboxyl group attacks the ortho position of the phenyl ring forming a five-membered ketone ring. The product S is 1-indanone (2,3-dihydro-1H-inden-1-one).

The combination Q = PhCH(Cl)COOH, R = PhCH${}_2$CH${}_2$COOH, S = 1-indanone matches option (B).

Correct option: (B).