= Solution
The hydroxylactam is a cyclic <hemiaminal>. Its alkoxide can open to the corresponding <aldehyde>/amidate, $\mathrm{MeNHCO-CH(OBn)-CH_2-CHO}$ in its protonated representation. This exposes an <aldehyde> without reducing or replacing the lactam <carbonyl>.
The ketophosphonate anion attacks that <aldehyde>. Formation and collapse of the phosphorus-containing intermediate give the predominantly E enone $\mathrm{MeNHCO-CH(OBn)-CH_2-CH=CH-COCH_3}$ and phosphate by <Horner-Wadsworth-Emmons reaction>. The amide nitrogen then closes by intramolecular <aza-Michael addition> at the enone beta carbon. <Protonation> of the <ketone> enolate gives the observed five-membered lactam with a $\mathrm{-CH_2COCH_3}$ substituent.
A folded, relatively unencumbered closure places OBn and the developing side chain in outward-facing positions; the expected major product has \b[OBn and the C5 acetonyl substituent cis]. With the OBn wedge retained, show the new C5–CH$_2$COCH$_3$ bond as a solid wedge. The closure is diastereoselective, while the original wavy <hemiaminal> centre is erased by <aldehyde> formation. This facial prediction rests on the preferred folded conformer rather than retention of either starting <hemiaminal> epimer.
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