@SamHopkins It seems at least to me to be unambiguous, since it is clear that there exists at least one $M$ that meets the informal description of enumerating and checking ZFC proofs, and Kleene's recursion theorem guarantees that it can encode the statement "M halts" in ZFC language

It's worth noting that if you take $\sum a_j s_j q_j$ (where $s_j, q_j$ are the dependent-on-$D$ polynomials, as described in the wiki article), instead of dividing by $p_j$ and taking the remainder, that you also similarly get a possibly higher degree polynomial that satisfies the congruences/matches the evaluations $E$.

wait...can it be that simple? I had just assumed the degree would be too large as $deg(\ell_j) \ge deg(a)$ plus then the $deg(a_j)$ increases it more. So then that sum may be interpolating some higher degree polynomial that agrees with $a$ on the domain. But it's not clear to me if it can be designed so cancellation of higher terms happens.

Is it obvious what is meant by "the Birkhoff decomposition" in this question? Because the decomposition of a doubly stochastic matrix into permutation matrices is not necessarily unique. So is your question "for all decompositions ..." or "does there exist a decomposition such that ..."?