With that said, even if the decision problem for ECDLP is in NP and P = NP, that does not mean that a solution to ECDLP is in P.
No, it's not. All instances of DLP, including ECDLP, are just asking for an x such that b^x = a, as Wikpedia will tell you [1]. They only differ in the choice of group, but exponentiation within the group is assumed to be an efficient operation.
> With that said, even if the decision problem for ECDLP is in NP, that does not mean that a solution to ECDLP is in P.
It trivially does. You simply start with an empty prefix x and extend it 1 bit at a time by repeatedly asking if (Curve_spec, P, x0) is in the language. If so you continue with x0, if not you continue with x1.
If you can solve that problem in P-time then you can also solve the functional version (find an x for which F(x) = 1) in P-time simply by setting each bit to a value for which the decision problem is satisfiable, one by one.