Consider the following reactions,

The compound [P] is:

Step 1: Analyze Reaction Sequence 1 (Diazotization and Azo Coupling)

$$\text{[P]} \xrightarrow{\text{(i) NaNO}_2/\text{HCl, } 0-5^\circ\text{C}} \text{Intermediate} \xrightarrow{\text{(ii) }\beta\text{-naphthol/NaOH}} \text{Colored Solid (Azo Dye)}$$

• The formation of a highly colored stable azo dye upon treatment with nitrous acid ($$\text{NaNO}_2/\text{HCl}$$) at low temperatures followed by coupling with $$\beta\text{-naphthol}$$ is a diagnostic test for primary aromatic amines ($$-\text{NH}_2$$ attached directly to a benzene ring).
• This eliminates N-methylaniline, which is a secondary amine and forms an insoluble yellow $N$-nitroso amine derivative instead of a diazonium salt.

Step 2: Analyze Reaction Sequence 2 (Exhaustive Bromination)

$$\text{[P]} \xrightarrow{\text{Br}_2/\text{H}_2\text{O}} \text{C}_7\text{H}_6\text{NBr}_3$$

• Let us work backward from the molecular formula of the brominated product, $$\text{C}_7\text{H}_6\text{NBr}_3$$.
• Aniline ($$\text{C}_6\text{H}_5\text{NH}_2$$) has 6 carbon atoms. Upon electrophilic substitution with excess bromine water ($$\text{Br}_2/\text{H}_2\text{O}$$), it yields 2,4,6-tribromoaniline, which contains 6 carbons ($$\text{C}_6\text{H}_4\text{NBr}_3$$).
• The target product formula contains 7 carbon atoms, which explicitly means a methyl group ($$-\text{CH}_3$$) must be present on the aromatic core along with the amino group. This implies that compound [P] is an isomer of toluidine (methylaniline).

Step 3: Pinpoint the Correct Toluidine Isomer ($$o\text{-}, \ m\text{-}, \text{ or } p\text{-}$$)

Let's calculate the formula of a toluidine isomer ($$\text{C}_7\text{H}_9\text{N}$$) undergoing a tribromination reaction: substituting 3 hydrogen atoms on the ring with 3 bromine atoms yields:

$$\text{C}_7\text{H}_9\text{N} + 3\text{Br}_2 \rightarrow \text{C}_7\text{H}_6\text{NBr}_3 + 3\text{HBr}$$

This matches our product formula perfectly. Now, let's see which isomer allows for exactly three positions to be available for bromination:

• In m-toluidine, the amino group ($$-\text{NH}_2$$) strongly activates positions 2, 4, and 6 (which are ortho and para to it). Crucially, none of these three highly activated positions are blocked by the methyl group (which resides at position 3). Thus, it undergoes smooth substitution at all three sites to form 2,4,6-tribromo-3-methylaniline.
• For o-toluidine or p-toluidine, one of these highly activated ortho/para positions would already be blocked by the methyl group, preventing a full tribromination.

Conclusion:

Combining the primary aromatic amine trait from the azo dye test with the open substitution positions required for a $$\text{C}_7\text{H}_6\text{NBr}_3$$ product confirms that [P] is m-toluidine.

Answer: Option C — m-toluidine