The reaction shows the halogenation of an amide group (\(\text{NHCOCH}_3\)) with bromine. Pyridine is a mild base and it directs the substitution at the position ortho to the existing \(\text{NHCOCH}_3\) group. The reaction produces an intermediate compound \(\text{NHCOCH}_3, \text{NH}_2\).
In the second step, bromine further reacts with the intermediate to give the final product.
Thus, the correct answer is \( \text{NHCOCH}_3, \text{NH}_2 \).