Zero Knowledge for Low Entropy Witness
$\newcommand{\NP}{\mathsf{NP}}\newcommand{\lang}{\mathcal{L}}\newcommand{\rel}{\mathcal{R}}\newcommand{\bin}{\{0,1\}}$
Let $\lang\subseteq\bin^*$ be any $\NP$ language with a corresponding $\NP$ relation $\rel$.
Now assume that every statement in $\lang$ has a short witness, i.e. a witness of constant (as in your question) or at most logarithmic length. I.e. it holds that
$$
  \forall\,x \in \lang\ \exists\,w\in\bin^{\log(|x|) }. (x,w)\in\rel
$$

By definition of an $\NP$ language, it is possible to check whether $(x,w)\in\rel$ in polynomial time for any pair $(x,w)$. Say it takes at most time $p(|x|)$ to decide. It thus follows that¹ we can give a polynomial time algorithm for deciding $\lang$ as follows:
\begin{align}
&amp;\underline{\mathcal{A}(x)\hspace{6em}}\\
&amp;\text{for } w\in\bin^{\log(|x|)}\\
&amp;\quad \text{if } (x,w)\in\rel \\
&amp;\quad\quad \text{return } 1\\
&amp;\text{return } 0
\end{align}

It is easy to see that this algorithm is deterministic, always decides correctly and runs in time at most $$2^{log(|x|)}\cdot p(|x|) = |x|\cdot p(|x|)$$ and therefore in polynomial time. We can thus conclude that any such language is in fact in $\mathsf{P}$.

And for any language there exists a trivial proof system, that consists of just sending nothing. The verifier can simply verify the statement themselves. Note that this proof system is trivially sound (as the prover cannot influence the verifier's decision at all), as well as trivially zero knowledge, since the simulator just needs to "simulate" the empty interaction.

Now you specifically asked about proofs of knowledge. It is easy to see that this is in fact also formally fulfilled here, as the algorithm $\mathcal{A}$ needs to be modified only slightly to extract $w$ from $x$ alone.

\begin{align}
&amp;\underline{\mathsf{Ext}(x)\hspace{6em}}\\
&amp;\text{for } w\in\bin^{\log(|x|)}\\
&amp;\quad \text{if } (x,w)\in\rel \\
&amp;\quad\quad \text{return } w\\
&amp;\text{return } \bot
\end{align}



¹As you in fact observed correctly.