For every $\mathrm{NP}$ language $L$, is there a verifier such that, for all the certificates $u$ of other verifiers of $L$, it accepts $(x, u)$? Let $L$ be an $\mathrm{NP}$ language.
Then there exists a verifier $V$ of $L$ and a polynomial $p\colon \mathbb{N} \to \mathbb{N}$, such that for every $x \in \Sigma^{*}$, $x \in L$ if and only if there exists a certificate $u \in \Sigma^{p(|x|)}$ for $x$ satisfying $V(x, u) = 1$.
Suppose that $\mathcal{V}_{q}$ be all the verifiers of $L$ such that the length of the certificates of $x$ is $q(|x|)$.
And for every $V \in \mathcal{V}_{q}$, let $V_{x}$ be all the certificates of $x$.
For every polynomial $q\colon \mathbb{N} \to \mathbb{N}$ satisfying $\mathcal{V}_{q} \neq \varnothing$, I want to know whether there is a verifier $\bar{V} \in \mathcal{V}_{q}$ such that for every $x \in L$,
$$\bar{V}_{x} = \bigcup_{V \in \mathcal{V}_{q}} V_{x}.$$
Fix $x \in L$ of length $|x|=n$. Suppose that $\mathcal{V}_q \neq \emptyset$, and let $m = q(n)$. Choose some $V^0 \in \mathcal{V}_q$. For any $s \in \{0,1\}^m$, let $V^s(y,u) = V^0(y,u \oplus s)$ for inputs of length $n$ and witnesses of length $m$, $V^s(y,u) = 0$ for inputs of length $n$ and witnesses lengths different from $m$, and $V^s(y,u) = V^0(y,u)$ otherwise. Then $V^s$ is a verifier for $L$ for all $s$, and so $\bigcup_{V \in \mathcal{V}_q} V_x = \{0,1\}^m$.
This means that if a verifier $\bar{V}$ exists, then $\bar{V}_x$ is the set of all strings of length $q(|x|)$. This means that $\bar{V}$ need not consult its witness, showing that $L \in \mathsf{P}$.