The major product 'P' for the following sequence of reactions is:

The reagents used in the four successive steps correspond to four very standard transformations in the chemistry of aromatic nitrogen-containing compounds. We identify each step first and then follow the fate of the functional group round by round.

Step 1 : Reduction of an aromatic nitro group
If the starting compound is $$C_6H_5NO_2$$ (nitrobenzene) and it is treated with a reducing system such as $$Sn/HCl$$, $$Fe/HCl$$ or $$H_2/Pd$$, the nitro group is converted into an amino group:$$C_6H_5NO_2 \;\xrightarrow[\;HCl\;]{\;Sn\;}\; C_6H_5NH_2$$.
Thus, after step 1 we obtain aniline.

Step 2 : Diazotisation of the amine
Primary aromatic amines react with nitrous acid (generated in situ from $$NaNO_2/HCl$$ at $$0{-}5^{\circ}\text{C}$$) to give diazonium salts:$$C_6H_5NH_2 \;\xrightarrow[\;0{-}5^{\circ}C\;]{\;NaNO_2/HCl\;}\; C_6H_5N_2^+Cl^-$$.

Step 3 : Sandmeyer cyanation
Aryl diazonium chloride reacts with cuprous cyanide to introduce the cyano group, giving an aromatic nitrile (Sandmeyer reaction):$$C_6H_5N_2^+Cl^- \;\xrightarrow{\;CuCN\;}\; C_6H_5CN$$.

Step 4 : Reduction of the nitrile to a primary amine
Lithium aluminium hydride $$\big(LiAlH_4\big)$$ is a powerful reducing agent that converts nitriles to primary amines by adding two hydrogen atoms to the carbon of the -CN group and two to the nitrogen:$$C_6H_5CN \;\xrightarrow{\;LiAlH_4\;}\; C_6H_5CH_2NH_2$$ (benzylamine).

Hence, after the entire sequence the major product $$P$$ is benzylamine, $$C_6H_5CH_2NH_2$$.

Among the given structures, benzylamine corresponds to Option B.

Therefore, the correct answer is:
Option B which is: benzylamine ($$C_6H_5CH_2NH_2$$).