= Solution
The useful object is the <Choi state> of the <entanglement-breaking channel>. Put $d_A=\dim\mathcal H_A$, introduce a reference $R$ with the same <computational basis> as $A$, and use the normalized <maximally entangled state>
$$
|\Phi\rangle_{RA}=d_A^{-1/2}\sum_j|j\rangle_R|j\rangle_A,\qquad
C_{RB}=(\operatorname{id}_R\otimes\mathcal E)(|\Phi\rangle\langle\Phi|).
$$
The <entanglement-breaking channel> makes $C_{RB}$ a <separable quantum state>. Its <partial trace> is $\operatorname{Tr}_B C_{RB}=I_R/d_A$, because $\mathcal E$ is a <quantum channel>. Write a <separable positive operator> decomposition
$$
C_{RB}=\sum_y\alpha_y\otimes\beta_y,\qquad \alpha_y\geq0,\quad\beta_y\geq0.
$$
Discard any term with $\operatorname{Tr}\beta_y=0$: a <positive operator> with zero <trace> is zero. Define
$$
\tau_y=\frac{\beta_y}{\operatorname{Tr}\beta_y},\qquad
E_y=d_A(\operatorname{Tr}\beta_y)\alpha_y^T.
$$
Each $\tau_y$ is a <density operator>. Each $E_y$ is a <positive operator>, and taking the <matrix transpose> of the <partial trace> identity gives
$$
\sum_y E_y=d_A\left(\sum_y(\operatorname{Tr}\beta_y)\alpha_y\right)^T=I_A.
$$
Thus the $E_y$ form a <POVM>. In this reference-first convention, the <Choi reconstruction formula> is
$$
\mathcal E(L)=d_A\operatorname{Tr}_R[(L^T\otimes I_B)C_{RB}]
=\sum_y\tau_y\operatorname{Tr}(E_yL).
$$
The factor $d_A$ is essential because we used a normalized <Choi state>. A <measurement channel> storing the outcome $y$, followed by the prescribed <quantum state preparation> of $\tau_y$, has precisely this action on every <linear operator> $L$. \b[The desired factorization is therefore]
$$
\boxed{\mathcal E^{B\leftarrow A}=\mathcal P^{B\leftarrow\widetilde Y}\mathcal M^{\widetilde Y\leftarrow A}.}
$$
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