Step-by-step Solution:

1. Analyze the Starting Materials (List I):

   • P: H-C≡N (Hydrogen cyanide). This is the simplest possible nitrile.
   • Q: CH₃-C≡N (Acetonitrile or ethanenitrile). This is a simple aliphatic nitrile.
   • R: C₆H₅-C≡N (Benzonitrile). This is an aromatic nitrile.
   • S: C₆H₅-N≡C (Phenyl isocyanide). This is an isocyanide, an isomer of benzonitrile.

2. Analyze the Reaction Schemes (List II):

   • I: (i) SnCl₂ + HCl, H₂O; (ii) H₃O⁺. These are the reagents for the Stephen reduction, a method to convert nitriles into aldehydes (R-C≡N → R-CHO).
   • II: (i) DIBAL-H, H₂O; (ii) H₃O⁺. Reduction with Diisobutylaluminium hydride (DIBAL-H) also partially reduces nitriles to aldehydes. It is a more modern and selective reagent than the one used in Stephen reduction.
   • III: (i) H₂O, NaOH, Boil. This represents alkaline hydrolysis. Nitriles (R-C≡N) are hydrolyzed to carboxylate salts (R-COO⁻Na⁺) and ammonia. Isocyanides (R-N≡C) are hydrolyzed to a primary amine (R-NH₂) and a formate salt (HCOO⁻Na⁺).
   • IV: (i) LiAlH₄, Ether. This is a strong reduction using Lithium aluminium hydride. Nitriles are completely reduced to primary amines (R-C≡N → R-CH₂NH₂). Isocyanides are reduced to secondary amines (R-N≡C → R-NH-CH₃).

3. Establish the Most Logical Matches: The best approach for matching questions is to identify the most specific or characteristic reactions first.

   • Matching R (C₆H₅-C≡N): Benzonitrile is an aromatic nitrile. A very important synthetic transformation is its conversion to benzaldehyde. DIBAL-H (Scheme II) is a highly effective and commonly used reagent for the partial reduction of nitriles (especially aromatic ones) to aldehydes. Therefore, R → 2 is an excellent match. $C_6H_5-C≡N \xrightarrow{(i) \text{ DIBAL-H, H}_2\text{O}} \xrightarrow{(ii) \text{ H}_3\text{O}^+} C_6H_5-CHO$

   • Matching Q (CH₃-C≡N): Acetonitrile is a simple aliphatic nitrile. A fundamental reaction is its complete reduction to the corresponding primary amine, ethylamine. Lithium aluminium hydride (LiAlH₄) (Scheme IV) is the classic strong reducing agent for this purpose. Therefore, Q → 4 is a very strong match. $CH_3-C≡N \xrightarrow{\text{LiAlH}_4, \text{ Ether}} CH_3-CH_2-NH_2$

   • Matching P (H-C≡N): Hydrogen cyanide is the simplest nitrile and can be considered the nitrile of formic acid. The most fundamental chemical relationship is its hydrolysis to a salt of formic acid (formate) under basic conditions (Scheme III). Thus, P → 3 is a very logical match. $H-C≡N \xrightarrow{\text{H}_2\text{O, NaOH, Boil}} HCOO^-Na^+ + NH_3$

4. Match the Remaining Pair by Elimination:

   • We have now established the matches: P→3, Q→4, and R→2.
   • The only remaining starting material is S (C₆H₅-N≡C), and the only remaining reaction scheme is I (Stephen reduction).
   • By elimination, S must match with 1.
   • Let's consider this reaction: C₆H₅-N≡C with SnCl₂/HCl followed by H₃O⁺. The acidic conditions (HCl, H₃O⁺) will cause the hydrolysis of the isocyanide to aniline and formic acid. C₆H₅-N≡C + 2H₂O → C₆H₅-NH₂ + HCOOH. Thus, a definite reaction occurs, making the pairing plausible within the context of the question.

5. Final Combination and Answer:

   • P is matched with 3.
   • Q is matched with 4.
   • R is matched with 2.
   • S is matched with 1.

   This combination is P-3, Q-4, R-2, S-1, which corresponds to option C.