Does a polynomial solution for an NP-complete problem that can only be implemented for small N *still* imply P=NP? Basic sanity check on NP-complete solutions:

Suppose there was a polynomial time solution for an NP-complete problem,
but the size of NP-complete problems that could be solved is still
relatively small (i.e. N = 32-64) due to limits in technology.

Would this show that P = NP? Or would this be more like the
pseudo-solution that the unary subset sum is in P?
If you mean that the polynomial-time algorithm only works for inputs up to some fixed size, it shows nothing at all. Any problem at all (even if it's undecidable, let alone NP-complete) becomes a finite language when restricted to instances of constant size. All finite languages can be decided in constant time.

If you mean that the polynomial-time algorithm works for all inputs but it's still so inefficient that current computers can only run it on small instances then, sure, that would prove that P$\,=\,$NP. The idea that "polynomial" means "efficiently and effectively solvable" is just a short-hand. The actual definitions of P and&nbsp;NP don't depend on that short-hand.