The correct match between Item-I (starting material) and Item-II (reagent) for the preparation of benzaldehyde is:
Item-I                            Item-II
(I) Benzene                   (P) HCl and $$\text{SnCl}_2$$, $$\text{H}_3\text{O}^+$$
(II) Benzonitrile            (Q) $$\text{H}_2$$, $$\text{Pd-BaSO}_4$$, S and quinoline
(III) Benzoyl Chloride    (R) Co, HCl and $$\text{AlCl}_3$$

We begin by recalling that to convert a given aromatic starting material into an aldehyde group $$\left(-CHO\right)$$, different classical name reactions are employed. The reagent sets themselves often reveal which reaction is intended. Hence, for every combination we shall first identify the reaction indicated by the reagent and then verify that the starting material is suitable for that reaction.

Let us look at Item I where the starting material is benzene $$\left(\text C_6\text H_6\right)$$. To introduce a formyl group directly on the benzene ring, the famous Gattermann-Koch formylation is used. The reagent for this reaction is stated, before any calculation, as

$$\text{CO} + \text{HCl} + \text{AlCl} _3\;($$ and sometimes CuCl $$).$$

This exactly matches reagent $$R : \text{CO},\;\text{HCl}\text{ and }\text{AlCl}_3.$$

Therefore, Item I must be paired with $$R.$$ After the formylation, the product obtained is benzaldehyde $$\left(\text C_6\text H_5\text{CHO}\right).$$

Next, consider Item II where the starting material is benzonitrile $$\left(\text C_6\text H_5\text{CN}\right).$$ Converting a nitrile into an aldehyde is carried out by the Stephen reduction. Before doing any algebra, the standard reagent for Stephen reduction is stated as

$$\text{SnCl} _2 + 2\, \text{HCl} \quad$$ followed by $$\quad \text{H} _3 \text{O} ^+.$$

This reagent set appears verbatim as $$P : \text{HCl and }\text{SnCl}_2,\;\text{H}_3\text O^+.$$

So, Item II must be paired with $$P.$$ The nitrile is thus reduced to the imidoyl chloride intermediate and on hydrolysis gives benzaldehyde.

Finally, Item III has benzoyl chloride $$\left(\text C_6\text H_5\text{COCl}\right)$$ as the starting material. Converting an acyl chloride to an aldehyde is the well-known Rosenmund reduction. The textbook reagent for Rosenmund reduction is stated as

$$\text H_2,\;\text{Pd-BaSO}_4$$ poisoned with sulfur or quinoline.

That is precisely reagent $$Q : \text H_2,\; \text{Pd-BaSO} _4,\; \text S\$$ and quinoline $$.$$

Hence, Item III must be paired with $$Q.$$ The catalytic hydrogenation stops at the aldehyde stage, giving benzaldehyde.

Summarising the matches:

$$\begin{aligned} \text{(I) Benzene} &\longrightarrow R \;(\text{Gattermann-Koch})\\ \text{(II) Benzonitrile} &\longrightarrow P \;(\text{Stephen reduction})\\ \text{(III) Benzoyl chloride} &\longrightarrow Q \;(\text{Rosenmund reduction}) \end{aligned}$$

The correspondence obtained is exactly

(I)-(R), (II)-(P) and (III)-(Q).

Looking at the options given, this set is listed as Option C.

Hence, the correct answer is Option C.