Products A and B formed in the following reactions are respectively:

The first step is the Hoffmann bromamide reaction.
Starting compound: $$C_6H_5CONH_2$$ (benzamide).
Reagents: $$Br_2/KOH$$ (basic medium).

The Hoffmann rearrangement removes one carbon atom from the amide and converts it into a primary amine that possesses one carbon fewer than the parent amide. Therefore
$$C_6H_5CONH_2 \;\xrightarrow{Br_2/KOH}\; C_6H_5NH_2$$

Hence product $$A$$ is $$C_6H_5NH_2$$ (aniline).

In the second step, product $$A$$ (aniline) is treated with nitrous acid generated in situ from $$NaNO_2/HCl$$ at $$0\!\!-\!\!5^{\circ}C$$. This forms a benzenediazonium chloride intermediate.
$$C_6H_5NH_2 + NaNO_2 + 2\,HCl \;\xrightarrow{0^\circ{\rm C}}\; C_6H_5N_2^+Cl^- + 2\,H_2O + NaCl$$

Benzenediazonium chloride is unstable in warm water; on gentle warming it undergoes hydrolysis to give phenol.
$$C_6H_5N_2^+Cl^- + H_2O \;\xrightarrow{\Delta}\; C_6H_5OH + N_2 \uparrow + HCl$$

Thus product $$B$$ is $$C_6H_5OH$$ (phenol).

Therefore, the two products are:
$$A : C_6H_5NH_2\;(\text{aniline}),\quad B : C_6H_5OH\;(\text{phenol})$$

Option A which is: aniline and phenol.