The major product of the following reaction is:

NaBH$$_4$$, MeOH, 25°C

First, the NaBH4 reduces the ketone to a secondary alcohol. This creates a halohydrin intermediate (a molecule with an -OH and a -Br on adjacent carbons).

In the presence of methanol or the slightly basic conditions generated during a borohydride reduction, the alcohol group can become deprotonated to form an alkoxide. This oxygen then performs an internal nucleophilic attack on the carbon holding the bromine.