= Solution
For a <memoryless quantum channel> $\mathcal N$, successive uses are described by $\mathcal N^{\otimes n}$. Its <entanglement-assisted classical capacity> is the supremum of classical bits per use achievable with error tending to zero when sender and receiver may use unlimited preshared <quantum entanglement>, independent of the message. Only the channel is used to transmit the message.
For input state $\rho_A$, choose a <purification> $|\psi_\rho\rangle_{RA}$ and set $\omega_{RB}=(\operatorname{id}_R\otimes\mathcal N)(|\psi_\rho\rangle\langle\psi_\rho|)$. The capacity formula is
$$
\boxed{C_E(\mathcal N)=\max_{\rho_A}I(R;B)_\omega
=\max_{\rho_A}\bigl[S(\rho_A)+S(\mathcal N(\rho_A))-S(\omega_{RB})\bigr].}
$$
<Purifications> differ only by an isometry on the reference, so this expression depends only on the input <density matrix>. The quantity optimized is <quantum mutual information>.
The binary <quantum erasure channel> sends a <qubit> to a <qubit> subspace plus an orthogonal erasure flag:
$$
\mathcal E_p(\rho)=(1-p)\rho\oplus p|e\rangle\langle e|,\qquad0\le p\le1.
$$
On the nonerased branch it transmits the entire state, including its coherences. Put $s=S(\rho)$. The <entropy of an orthogonal quantum mixture> gives
$$
S(\mathcal E_p(\rho))=h_2(p)+(1-p)s.
$$
The reference-output state has two orthogonal branches:
$$
\omega_{RB}=(1-p)|\psi_\rho\rangle\langle\psi_\rho|\ \oplus\ p\rho_R\otimes|e\rangle\langle e|.
$$
The first branch is pure; the second has entropy $S(\rho_R)=s$. Thus $S(\omega_{RB})=h_2(p)+ps$ and
$$
I(R;B)_\omega=s+h_2(p)+(1-p)s-h_2(p)-ps=2(1-p)s.
$$
A <qubit> has entropy at most one bit, with equality at $\rho=I_2/2$. We conclude
$$
\boxed{C_E(\mathcal E_p)=2(1-p)\text{ bits per channel use}.}
$$
The endpoint capacities are two and zero. As an operational check, <superdense coding> encodes four classical messages in one member of a shared Bell pair. A successful channel use lets the receiver identify the message; an erased <qubit> yields the flag. This induces a four-symbol classical erasure channel of capacity $(1-p)\log_24=2(1-p)$, agreeing with the entropy calculation.
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