Does sodium borohydride reduce C=N group? Q.15 the major product of the following reaction is:


I have read that $\ce{NaBH4}$ reduces only aldehyde, ketone and acid chlorides. In some cases it may reduce double bonds conjugated with carbonyl groups. So shouldn't the answer here be option 1?
Waylander's comments answer the title question as well as I could.  Sodium borohydride, and even modified versions where some of the hydride ligands are replaced by others like $\ce{NaBH3CN}$, can reduce imines.

But, if we have only a limited amount of borohydride we will likely see the carboxyl group primarily reduced, as it has a more polar pi bond and thus it is a more reactive electrophile.  So you may in principle get mostly 1 with limited sodium borohydride, or 4 if you have enough sodium borohydride to attack both the carboxyl and less reactive imine functions.

The carbon-carbon bond here, not conjugated to either electronegatove function, is interactive to borohydride and thus products 2 and 3 are not reached.