Consider the above reaction. The product A and product B respectively are

Consider the above reaction, the product and product respectively are

The starting compound is aniline, $$C_6H_5NH_2$$.

Step-1 : Diazotisation (formation of A)
When aniline is treated with sodium nitrite in the presence of dilute hydrochloric acid at $$0\!-\!5^{\circ}\text{C}$$, it undergoes diazotisation. The nitrous acid generated in situ converts the -NH2 group into a diazonium group.

$$C_6H_5NH_2 + NaNO_2 + 2\,HCl \;\xrightarrow[0-5^{\circ}\!C]{}\; C_6H_5N_2^+Cl^- + NaCl + 2\,H_2O$$

Thus, product $$A$$ is benzene diazonium chloride, $$C_6H_5N_2^+Cl^-$$.

Step-2 : Sandmeyer reaction (formation of B)
A diazonium salt reacts with cuprous cyanide (generated from $$CuCN/KCN$$) to give the corresponding aryl nitrile. This transformation is called the Sandmeyer cyanation.

$$C_6H_5N_2^+Cl^- + CuCN \;\xrightarrow{\,50-60^{\circ}\!C\,}\; C_6H_5CN + CuCl + N_2\uparrow$$

Therefore, product $$B$$ is benzonitrile, $$C_6H_5CN$$.

Hence, the pair of products is:
$$A : C_6H_5N_2^+Cl^- \quad\text{and}\quad B : C_6H_5CN$$

These correspond to Option A.