This is a matching question where we need to identify the major products for each reaction given in LIST-I from the options in LIST-II.

Step-by-step analysis:

1. Reaction P: Anisole + $CH_3Cl$ / Anhyd. $AlCl_3$

   • This is a Friedel-Crafts alkylation reaction. The methoxy group ($-OCH_3$) on anisole is an activating group and directs the incoming electrophile ($CH_3^+$) to the ortho and para positions.
   • The main products are p-methylanisole (4-methoxytoluene) and o-methylanisole (2-methoxytoluene).
   • p-methylanisole corresponds to product (4) and o-methylanisole corresponds to product (5) in LIST-II. The para product (4) is typically the major isomer due to less steric hindrance.
   • Additionally, anhydrous $AlCl_3$ can act as a Lewis acid to cleave the ether bond in anisole, especially at elevated temperatures, forming phenol. This phenol can then undergo Friedel-Crafts alkylation to yield o-cresol (product (1)) and p-cresol.
   • Thus, products (1), (4), and (5) are all possible products for reaction P.

2. Reaction Q: Anisole + Styrene ($C_6H_5CH=CH_2$) / $H^+$

   • In the presence of an acid catalyst ($H^+$), styrene gets protonated to form a stable secondary benzylic carbocation ($C_6H_5\stackrel{+}{C}H-CH_3$).
   • This carbocation acts as an electrophile and attacks the electron-rich anisole ring (electrophilic aromatic substitution).
   • The $-OCH_3$ group directs the substitution to the ortho and para positions. Due to the bulky nature of the electrophile, the para product is formed as the major product.
   • The para product is 1-methoxy-4-(1-phenylethyl)benzene, which is product (2) in LIST-II.
   • Therefore, Q → 2.

3. Reaction S: Anisole + HI(conc.) / heat

   • This reaction involves the cleavage of the ether linkage in anisole by a strong nucleophilic acid, concentrated HI. This is known as the Zeisel method.
   • The reaction proceeds via nucleophilic attack of $I^-$ on the methyl group ($S_N2$ mechanism), as the phenyl-oxygen bond has partial double bond character and is stronger.
   • The products are phenol and methyl iodide ($CH_3I$). $C_6H_5-O-CH_3 + HI \xrightarrow{\Delta} C_6H_5-OH + CH_3I$
   • Phenol corresponds to product (3) in LIST-II.
   • Therefore, S → 3.

4. Evaluating the Options based on Q and S

   • Our analysis shows that Q → 2 and S → 3. Let's examine the given options:
     • A: $P - 1,5; Q - 2; R - 3; S - 4$ (Incorrect for S)
     • B: $P - 1,4; Q - 2; R - 4; S - 3$ (Correct for Q and S)
     • C: $P - 1,4; Q - 1,2; R - 3,4; S - 4$ (Incorrect for Q and S)
     • D: $P - 4,5; Q - 4; R - 4; S - 3,4$ (Incorrect for Q and S)
   • Only option B has the correct matching for both Q and S.

5. Final Check of Option B

   • P → 1, 4: This suggests the products are o-cresol (1) and p-methylanisole (4). As explained in step 1, this is plausible, with (4) being a main product and (1) being a side product. This match is acceptable.
   • R → 4: Reaction R is the Fries rearrangement of phenyl ethanoate ($CH_3COOC_6H_5$) with anhydrous $AlCl_3$ and heat. The expected products are o-hydroxyacetophenone and p-hydroxyacetophenone, which are not in LIST-II. The transformation of phenyl ethanoate to p-methylanisole (4) is not a standard or plausible reaction under these conditions. There appears to be an error in the question for this specific match.
   • However, since Q → 2 and S → 3 are definitively correct and only option B aligns with these facts, it is the most logical choice among the given options, despite the apparent error in the R → 4 mapping.

Conclusion: Based on the process of elimination using the correct matches for Q and S, option B is the intended answer.