The final product C, obtained in this reaction, would be :

Step 1 (A): acetic anhydride acetylates the $-NH_2$ group of p-toluidine, giving p-methylacetanilide, $p-CH_3C_6H_4NHCOCH_3$. This protects the amino group and reduces its strong activating effect, so that only monobromination occurs.
Step 2 (B): bromine in acetic acid substitutes on the ring. The $-NHCOCH_3$ group is ortho/para directing; the para position is occupied by $CH_3$, so Br enters ortho to $-NHCOCH_3$, giving 2-bromo-4-methylacetanilide.
Step 3 (C): acid hydrolysis removes the acetyl group and frees the amino group:
$2-Br-4-CH_3C_6H_3NHCOCH_3 + H_2O \xrightarrow{H^+} 2-Br-4-CH_3C_6H_3NH_2 + CH_3COOH$
So C is 2-bromo-4-methylaniline: $NH_2$ on the ring, Br ortho to it and $CH_3$ para to it. (B, the acetamide, is the intermediate before hydrolysis.)