Reaction of iso-propylbenzene with $$O_2$$ followed by the treatment with $$H_3O^+$$ forms phenol and a by-product P. Reaction of P with 3 equivalents of $$Cl_2$$ gives compound Q. Treatment of Q with $$Ca(OH)_2$$ produces compound R and calcium salt S.

The correct statement(s) regarding P, Q, R and S is(are)

The cumene (Hock) process converts iso-propylbenzene (cumene) to phenol and acetone by way of the cumene-hydroperoxide rearrangement.
Hence, in the first step the by-product is

$$P = (CH_3)_2C=O \;(\text{acetone}).$$

Preparation of Q
Each molecule of $$Cl_2$$ supplies two chlorine atoms. Three “equivalents” (three moles) of $$Cl_2$$ therefore furnish six chlorine atoms - exactly the number needed to replace all six α-hydrogens of acetone. Chlorination in sunlight thus converts both methyl groups into trichloromethyl groups:

$$CH_3COCH_3 \;\xrightarrow[\;h\nu\;]{3\,Cl_2}\; CCl_3COCCl_3$$

so

$$Q = CCl_3COCCl_3\;(\text{hexachloroacetone}).$$

Formation of R and S
A 1,1,1-trihalo ketone undergoes the haloform reaction even in simple alkali. Treating Q with slaked lime gives one molecule of chloroform (haloform) together with the calcium salt of the residual acid:

$$CCl_3COCCl_3 + 2\,Ca(OH)_2 \;\longrightarrow\; 2\,COCl_3^-Ca^{2+} + 2\,CHCl_3 + 2\,H_2O$$

Thus

$$R = CHCl_3\;(\text{chloroform}),\qquad S = Ca(CCl_3COO)_2\;(\text{calcium trichloroacetate}).$$

Checking the statements

Option A Acetone (P) condenses with chloroform (R) in strong base (KOH) to give chlorobutanol (also called chloretone). Acidification merely neutralises the medium: the product remains unchanged. Hence statement A is correct.

Option B For the slow photo-oxidation of chloroform moisture is essential; dry $$CHCl_3$$ exposed only to $$O_2$$ and light does not accumulate phosgene. Therefore statement B is not accepted as correct for the conditions written.

Option C Alkaline hydrolysis of Q liberates chloroform and the trichloroacetate ion; it never produces $$Cl_3CCH_2OH$$. Hence statement C is incorrect.

Option D On heating, calcium trichloroacetate (S) does not regenerate acetone. Therefore statement D is incorrect.

Only statement A is correct.

Final answer: Option A which is: Reaction of P with R in the presence of KOH followed by acidification gives chlorobutanol.