List-I contains various reaction sequences and List-II contains the possible products. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.

List-I		List-II
(P)		(1)
(Q)		(2)
(R)		(3)
(S)		(4)
		(5)

First write down the complete data of the question.

List-I (reaction sequences):
P. $$C_6H_5OCH_3 \;\xrightarrow[\text{excess}]{\,HI\,}$$ (anisole is heated with excess hydroiodic acid)
Q. $$C_6H_5OH \;\xrightarrow[]{\,conc.\;HNO_3/\,conc.\;H_2SO_4\,}$$ (phenol treated with conc. nitrating mixture)
R. $$C_6H_5OH \;\xrightarrow[]{\,CH_3COCl/pyridine\,}$$ (phenol acetylated with acetyl chloride in pyridine)
S. $$C_6H_5OH \;\xrightarrow[]{\,NaOH,\;140^{\circ}C,\;7\,atm\,}\;NaO\!C_6H_4O^-$$ $$\xrightarrow[]{\,CO_2\,}\xrightarrow[]{\,H^+\,}$$ (Kolbe-Schmitt reaction followed by acidification)

List-II (probable products):
1. $$o\text{-}HO\!C_6H_4COOH$$  (salicylic acid)
2. $$p\text{-}HO\!C_6H_4NO_2$$  (p-nitrophenol)
3. $$C_6H_5OH + CH_3I$$
4. $$C_6H_5OCOCH_3$$  (phenyl acetate)
5. $$2,4,6\text{-}trinitrophenol$$  (picric acid)

Now analyse every reaction one by one.

Anisole $$\left(C_6H_5OCH_3\right)$$ when heated with excess $$HI$$ undergoes cleavage of the C-O bond attached to the alkyl group, forming phenol and methyl iodide: $$C_6H_5OCH_3 + HI \;\longrightarrow\; C_6H_5OH + CH_3I$$ This matches product 3.

Phenol reacts with concentrated $$HNO_3/H_2SO_4$$ at low temperature to give picric acid (2,4,6-trinitrophenol) because -OH is a strong activator and three nitro groups finally enter the ring: $$C_6H_5OH \xrightarrow[]{conc.\;HNO_3/H_2SO_4} 2,4,6\text{-}trinitrophenol$$ Hence Q corresponds to product 5.

Acetylation of phenol with acetyl chloride in pyridine gives phenyl acetate: $$C_6H_5OH + CH_3COCl \;\xrightarrow[]{pyridine}\; C_6H_5OCOCH_3 + HCl$$ Therefore R matches product 4.

The Kolbe-Schmitt reaction converts phenoxide ion to o-hydroxybenzoate ion which after acidification yields salicylic acid: $$C_6H_5OH \xrightarrow[]{NaOH} C_6H_5O^-Na^+ \xrightarrow[7\,atm]{CO_2,\;140^{\circ}C} o\text{-}HO\!C_6H_4COONa \xrightarrow[]{H^+} o\text{-}HO\!C_6H_4COOH$$ So S corresponds to product 1.

Collecting all matches:
P → 3, Q → 5, R → 4, S → 1

The option that contains this combination is Option A.

Final answer: Option A which is: P-3, Q-5, R-4, S-1.