Consider the above reaction, the Product "P" is:

Nitrobenzene undergoes reduction with:

$$\mathrm{Sn/HCl}$$

The nitro group is completely reduced to a primary amine.

$$\mathrm{C_6H_5NO_2 \xrightarrow{Sn/HCl} C_6H_5NH_2}$$

Thus, intermediate $$\mathrm{A}$$ is:

$$\mathrm{Aniline\ (C_6H_5NH_2)}$$

In the next step, aniline reacts with benzenediazonium chloride:

$$\mathrm{C_6H_5N_2^+Cl^-}$$ through azo coupling.

The diazonium ion acts as a weak electrophile, while aniline activates the benzene ring through the:

$$\mathrm{-NH_2}$$ group. Since: $$\mathrm{-NH_2}$$ is ortho/para directing, coupling occurs mainly at the para position due to lower steric hindrance.

The reaction is:

$$\mathrm{C_6H_5NH_2 + C_6H_5N_2^+Cl^- \longrightarrow p\text{-}H_2NC_6H_4-N=N-C_6H_5 + HCl}$$

The final product $$\mathrm{P}$$ is:

$$\mathrm{p\text{-}aminoazobenzene}$$

which is a yellow azo dye.