= Solution
Write $X$ and $Y$ for the classical <quantum registers> $\widetilde X$ and $\widetilde Y$. By the <quantum mutual information> formula $I(U:V)=S(U)+S(V)-S(UV)$, expanding the right-hand side of the <quantum mutual information balance identity> gives
$$
\begin{aligned}
&I(X:B)+I(XB:D)-I(B:D)\\
&=S(X)+S(B)-S(XB)+S(XB)+S(D)-S(XBD)\\
&\qquad-S(B)-S(D)+S(BD)\\
&=S(X)+S(BD)-S(XBD)=I(X:BD).
\end{aligned}
$$
All <Von Neumann entropies> here are evaluated in $\omega$.
We use three facts: <quantum mutual information> is nonnegative by <nonnegativity of quantum relative entropy>; a local <quantum channel> cannot increase it by <data processing for quantum mutual information>; and the <quantum mutual information> of a <classical-quantum state> is its ensemble's <Holevo quantity>. Write $\chi(\mathcal T)$ for the <Holevo capacity>, the supremum of the output <Holevo quantity> over finite input ensembles.
The $XB$ marginal is an output ensemble for $\mathcal E$, with inputs $\rho(x)_A=\operatorname{Tr}_C\rho(x)_{AC}$. Consequently $I(X:B)_\omega\leq\chi(\mathcal E)$. Also $\omega_{XBD}$ is obtained from $\sigma_{XYD}$ by a <quantum channel> on $XY$ which retains $X$ and prepares $B$ from $Y$. Hence
$$
I(XB:D)_\omega\leq I(XY:D)_\sigma.
$$
To bound the latter even for <entangled states> $\rho(x)_{AC}$, exhibit the <conditional input ensemble after a local measurement>. Set
$$
q(y|x)=\operatorname{Tr}[(E_y\otimes I_C)\rho(x)_{AC}],\qquad
\rho_C^{x,y}=\frac{\operatorname{Tr}_A[(\sqrt{E_y}\otimes I_C)\rho(x)_{AC}(\sqrt{E_y}\otimes I_C)]}{q(y|x)}
$$
when $q(y|x)>0$; zero-weight outcomes can be omitted. The numerator is a <positive operator>, its <trace> is $q(y|x)$, and the $q(y|x)$ sum to one. Thus
$$
\sigma_{XYD}=\sum_{x,y}P_X(x)q(y|x)|x,y\rangle\langle x,y|\otimes\mathcal N(\rho_C^{x,y}).
$$
This is a <classical-quantum state> with an output ensemble for $\mathcal N$, so $I(XY:D)_\sigma\leq\chi(\mathcal N)$. Combining these bounds with $I(B:D)_\omega\geq0$ yields
$$
\boxed{I(X:BD)_\omega\leq\chi(\mathcal E)+\chi(\mathcal N),\qquad
\chi(\mathcal E\otimes\mathcal N)\leq\chi(\mathcal E)+\chi(\mathcal N).}
$$
The last step takes the supremum over all input ensembles on $AC$. \b[The left-hand channel direction is $B\leftarrow A$], as established in part (i); the printed $A\leftarrow B$ in the last inequality is a typographical reversal. Independent product ensembles also give the reverse inequality, so this proves <Holevo-capacity additivity for entanglement-breaking channels>.
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