= Solution
Apply the <measure-and-prepare channel> locally to an arbitrary bipartite <density operator> $\rho_{AR}$. The resulting state is
$$
(\Lambda\otimes\operatorname{id}_R)(\rho_{AR})=\sum_a\sigma_a\otimes R_a,
$$
where
$$
R_a=\operatorname{Tr}_A\left[(\sqrt{E_a}\otimes I)\rho_{AR}(\sqrt{E_a}\otimes I)\right]\geq0.
$$
To see that this is the correct output, the <partial trace> is cyclic for operators acting only on the traced subsystem, so the displayed expression equals $\operatorname{Tr}_A[(E_a\otimes I)\rho_{AR}]$. Put $q_a=\operatorname{Tr}R_a$. The <POVM> completeness relation implies $\sum_aq_a=1$, and terms with $q_a=0$ vanish. Therefore
$$
(\Lambda\otimes\operatorname{id}_R)(\rho_{AR})
=\sum_{a:q_a>0}q_a\sigma_a\otimes\frac{R_a}{q_a}
$$
is a <separable quantum state>. Thus the channel is an <entanglement-breaking channel>, for every reference-system dimension.
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