I'd say the closest thing to a "classical channel" corresponding to a conditional probability distribution $p(y|x)$ is a quantum channel of the form
$$\Phi(\rho) = \sum_{a,b} \langle a|\rho|a\rangle p(b|a) \, |b\rangle\!\langle b| \equiv \sum_a \langle \Pi_a,\rho\rangle \sigma_a, \quad \sigma_a\equiv\sum_b p(b|a) \Pi_b,$$
where $\Pi_a\equiv |a\rangle\!\langle a|$.
This is the channel that, conditionally to the input state being in the $a$-th state (when measured in an appropriate basis), gives the outcome $b$ (again, wrt some choice of measurement basis) with probability $p(b|a)$.
You can notice this is a special case of an entanglement-breaking channel. As mentioned e.g. in this answer, these have Choi equal to
$$J(\Phi) = \sum_a\sigma_a\otimes \Pi_a^T.$$
Just as an addendum, one effectively considers similar channels e.g. when discussing Holevo's bound. In there, one is interested in figuring out how much of the (classical) information encoded in a bunch of quantum states $\sigma_a$ can be later recovered through some measurement $\mu_b$ on said states. One thus wonders about the mutual information of the state
$$\sum_{a,b} \langle \mu_b,\sigma_a\rangle\, ( \Pi_a\otimes \Pi_b).$$
Notice the similarity with the $\Phi$ in the first equation.
