= Solution
Write a qubit <density operator> in <Bloch vector> form, $\rho=(I+\mathbf r\mathbin\cdot\boldsymbol\sigma)/2$, with $|\mathbf r|\leq1$. The <quantum depolarizing channel> maps $\mathbf r$ to $p\mathbf r$, so its output eigenvalues are $(1\pm p|\mathbf r|)/2$. The output <Von Neumann entropy> is minimized on pure inputs, where it equals $h_2((1+p)/2)$.
For any input ensemble, its output <Holevo quantity> is at most the maximum qubit entropy minus this minimum output entropy:
$$
\chi\leq1-h_2\left(\frac{1+p}2\right).
$$
Equality is attained by two equiprobable orthogonal pure inputs: their average output is $I/2$, while each output has the minimum entropy. Thus
$$
\boxed{\chi^*(\Lambda_{\rm dep})=1-h_2\left(\frac{1+p}2\right)}.
$$
This uses the question's retention parameter $p$; in the usual convex-mixture range, $0\leq p\leq1$. The expression also holds throughout the full completely positive qubit range $-1/3\leq p\leq1$, since $h_2((1+p)/2)=h_2((1+|p|)/2)$.
The supplied <additivity of Holevo capacity> gives $\chi^*(\Lambda_{\rm dep}^{\otimes n})=n\chi^*(\Lambda_{\rm dep})$. The <Holevo-Schumacher-Westmoreland theorem> expresses the unassisted <classical capacity of a quantum channel> as the regularized <Holevo capacity>, so
$$
\boxed{C(\Lambda_{\rm dep})=\chi^*(\Lambda_{\rm dep})}.
$$
Therefore entangled inputs across channel uses cannot increase this classical capacity.
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