= Solution
Use the normalized <Choi state> $C_\Lambda=(\Lambda\otimes\operatorname{id})(|\Phi_d\rangle\langle\Phi_d|)$, with $|\Phi_d\rangle=d^{-1/2}\sum_i|ii\rangle$. If $\Lambda$ is an <entanglement-breaking channel>, its action on this particular bipartite input makes $C_\Lambda$ a <separable quantum state>.
Conversely, suppose $C_\Lambda=\sum_ap_a\sigma_a\otimes\tau_a$, with local <density operators> $\sigma_a,\tau_a$. The <Choi reconstruction formula> gives
$$
\Lambda(X)=d\operatorname{Tr}_B\left[C_\Lambda(I\otimes X^T)\right]
=\sum_a\operatorname{Tr}(E_aX)\sigma_a,\qquad E_a=dp_a\tau_a^T\geq0.
$$
Because $\Lambda$ is trace preserving, $\operatorname{Tr}_A C_\Lambda=I/d$, so $\sum_aE_a=I$. Thus the $E_a$ form a <POVM> and the channel is a <measure-and-prepare channel>.
For any bipartite input $\rho_{AR}$, define the positive, possibly unnormalized reference operators
$$
R_a=\operatorname{Tr}_A\left[(\sqrt{E_a}\otimes I)\rho_{AR}(\sqrt{E_a}\otimes I)\right].
$$
Its output is $\sum_a\sigma_a\otimes R_a$. Since $\sum_a\operatorname{Tr}R_a=1$, normalizing each nonzero $R_a$ expresses this as a <convex combination> of <product states>. Hence every output is separable, proving the <separable Choi-state criterion for entanglement breaking>.
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