 Solution

B is 4-hydroxy-1-methylpyrrolidin-2-one; C is 1-methylpyrrolidin-3-ol. Their ring connectivities are shown below. Each ring remains five membered; the attached N-methyl group is retained.

For B, a B–H bond of sodium borohydride supplies a hydride equivalent to the ketone carbonyl carbon, and the carbonyl $\pi$ pair moves to oxygen. This gives an alkoxide, whose protonation during the reaction or workup gives the secondary alcohol. The lactam carbonyl remains. Converting one $\mathrm{C}=\mathrm{O}$ to $\mathrm{CH}(\mathrm{OH})$ adds two H atoms and gives ${\mathrm{C}}_5{\mathrm{H}}_9\mathrm{N}{\mathrm{O}}_2$.

For C, lithium aluminium hydride reduces that ketone in the same net way and also reduces the lactam. Coordination of the amide oxygen to aluminium assists hydride attack at its carbonyl carbon, giving a tetrahedral intermediate. The N lone pair then forms a C=N bond while the C–O bond breaks to an aluminium-bound oxygen group. A resulting iminium ion accepts a second hydride at carbon, yielding the cyclic tertiary amine. Aqueous workup protonates the ketone-derived alkoxide.

The lactam conversion replaces $\mathrm{N}-\mathrm{C}(=\mathrm{O})$ by $\mathrm{N}-\mathrm{C}{\mathrm{H}}_2$; its oxygen is removed rather than retained as another alcohol. Together with ketone reduction, the net change is $+4\mathrm{H}-\mathrm{O}$, giving ${\mathrm{C}}_5{\mathrm{H}}_{11}\mathrm{NO}$. No stereochemistry is specified: the new alcohol stereocentre is formed as a racemic mixture in an achiral environment.