= Solution
The carbon–phosphorus coupling in the $^{13}$C spectrum is consistent with deprotonation of the methyl group rather than loss of that group:
$$
\boxed{\mathrm{A=Ph_2P(O)CH_2Li}.}
$$
Addition to the nitrile initially gives $\mathrm{Ph_2P(O)CH_2C(tBu)=NLi}$. An intramolecular <proton transfer> then converts the imidoyl anion into a conjugated <lithiated enamide>:
$$
\boxed{\mathrm{B=[Li\{N(H)C(tBu)=CHP(O)Ph_2\}]_4}.}
$$
The phosphorus–carbon bond is retained. \b[The rearrangement is a hydrogen migration, not a migration of the phosphoryl group to nitrogen.] Addition of half an equivalent of <TMEDA> per Li gives the hemisolvated dimer
$$
\boxed{\mathrm{C=[Li_2\{N(H)C(tBu)=CHP(O)Ph_2\}_2(TMEDA)]}.}
$$
The tetrameric and dimeric nuclearities are established for this reaction system; the spectra in the question are consistent with them, but spectra alone would not uniquely prove every detail of the aggregates.<Lithiated organophosphorus enamines: a new synthetic approach and the first crystal structures by William Clegg and others>[Abstract and structures of the unsolvated enamide and its TMEDA hemisolvate, pages 1401–1402]{ref}
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