Question about NP problem certificates and P=NP From my understanding a problem is considered to be in NP time if it can be solved in polynomial time with a non-deterministic Turing machine and verified in polynomial time with a certificate.

My question is whether or not creating an algorithm to determine values accepted by a certificate for an NP problem can itself be considered an NP problem.

This problem is an example of what I had in mind (it's more or less a variation of SAT).

Imagine a padlock with N switches and each switch has a configuration 'UP' or 'DOWN'. The padlock can have anywhere between 1 and $2^N$ solutions (i.e. it may only accept one combination or any number of combinations up to and including all possible combinations) and both the total number of solutions and the solutions themselves are selected at random.

If there are $2^N$ solutions a solution takes $O(1)$ time to find since all inputs are valid.

However, if there is only 1 solution $O(2^N)$ time is required to find it.

If it is accepted that trying all combinations on the padlock requires $2^N$ combinations and that because the solution is random there is no way to guarantee a solution unless $2^N - 1$ combinations are tested by process of elimination.

Because the total number of acceptable solutions is not known until different candidates are tested against the certificate, the problem requires $O(2^N)$ time to solve.

To prove that the problem is solvable in NP, use an NTM to solve the problem by non-deterministically testing all valid combinations and accepting any valid solution. Takes $O(N)$ time to solve since every differing possible bit to test will branch into a new non-deterministic chain.

The certificate for the problem contains acceptable values for each 'swicth' (bit) at each location either 1 for 'UP', 0 for 'DOWN' or 'X' to accept any value at that location and takes $O(N)$ time to verify.

Is this valid?

If I'm correct and the problem above does fit in the class of NP it seems sufficient to prove that $P \neq NP$ but I'm fairly confident I made an error I'm not aware of and I'm not sure if my example can be properly classed as an NP problem.
Your problem is in the black box or oracle setting, that is, you are given access to the input as a black box rather than explicitly. You can contrast your problem with 3SAT, in which the input is a formula. One way to solve 3SAT is to try plugging in solutions at random (or systematically) and checking whether they satisfy the formula. However, better approaches are known, and these approaches use the explicit description of the formula.

The P vs NP question is in the white box setting – concretely, $\mathsf{P} \neq \mathsf{NP}$ is equivalent to the conjecture that no polynomial time algorithm solves 3SAT. Results in black box settings are irrelevant (at least directly) to the P vs NP question.

In the black box model we can also separate P from BPP (at least for promise problems), by considering the problem of finding a zero input for a balanced Boolean function. However, for explicit inputs, it is considered likely that P=BPP. This shows that the black box setting exhibits different behavior than the white box setting.

Black box settings are common in optimization, where often we are looking for algorithms which do not rely on the exact form of the input function. For example, in submodular optimization we are given a submodular function as a black box. One method for proving lower bounds in this area is the indistinguishability method, in which we construct two black boxes which are identical on most inputs but have a very different optimal solution. This method is technically much easier than similar results in the white box setting (for example, in the special case of set coverage functions or cut functions), which rely on PCP theory.