For every $\mathrm{NP}$ language $L$, is there a verifier such that, for all the certificates $u$ of other verifiers of $L$, it accepts $(x, u)$?
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 $L \in \mathsf{P}$.