Monobromination of napthalene derivative
You have electronically asymmetric rings. The right ring is deactivated by the carbonyl group that acts like an acyl group (remember that acyl groups are strong meta-deactivators). The left ring is activated by both $\ce{-CH3}$ and the nitrogen substituent acting as an amine (activating ortho/para directing), so bromination will occur on the left ring (since it is more nucleophilic).

Monobromination is not guaranteed experimentally. Although the right ring is deactivated under the right condition (at higher temperatures) you can end up with more than one $\ce{Br}$ being added. Maybe this question is just showing that a bromination will occur probably in that particular position.