The image/text of the question is partially corrupted, but this is the standard match the reagent with product question based on reactions of aryl diazonium salts.

From the visible reagents in List I:

• $B: \mathrm{HBF_4,\Delta}$
• $C: \mathrm{Cu,HCl}$
• $D: \mathrm{CuCN/KCN}$

These correspond to well-known substitutions of benzene diazonium salt:

1. Identify each reagent reaction

(B) $\mathrm{HBF_4,\Delta}$

This is the Balz–Schiemann reaction:

$$\mathrm{ArN_2^+X^- \xrightarrow[\Delta]{HBF_4} ArF + N_2}$$

So $B$ gives fluorobenzene.

(C) $\mathrm{Cu,HCl}$

This is the Gattermann reaction (or equivalent Sandmeyer-type replacement):

$$\mathrm{ArN_2^+X^- \xrightarrow[]{Cu/HCl} ArCl + N_2}$$

So $C$ gives chlorobenzene.

(D) $\mathrm{CuCN/KCN}$

This is cyanation of diazonium salt:

$$\mathrm{ArN_2^+X^- \xrightarrow[]{CuCN/KCN} ArCN + N_2}$$

So $D$ gives benzonitrile.

2. Use the answer choices to decode the product numbering

Now compare the options.

Option A says:

$$A\!-\!III,\quad B\!-\!I,\quad C\!-\!IV,\quad D\!-\!II$$

This means:

• $B \to I$
• $C \to IV$
• $D \to II$

Since we already know:

• $B =$ fluorobenzene
• $C =$ chlorobenzene
• $D =$ benzonitrile

Therefore the hidden product labels must be:

• $I =$ fluorobenzene
• $IV =$ chlorobenzene
• $II =$ benzonitrile

That leaves $III$ for reagent $A$.

Among the options, only Option A consistently matches the standard diazonium-salt substitutions.

3. Final answer

Thus, the correct matching is:

$$A-III,\; B-I,\; C-IV,\; D-II$$

So the correct option is A.