= Solution
Choose <Kraus representations>
$$
\mathcal C(L)=\sum_a C_aLC_a^\dagger,\qquad
\mathcal D(M)=\sum_b D_bMD_b^\dagger,\qquad
\sum_aC_a^\dagger C_a=I_Q,\quad\sum_bD_b^\dagger D_b=I_K.
$$
The composite <quantum channel> has <Kraus operators> $T_{ba}=D_bC_a$. Each has <matrix rank> at most $k$, because it factors through the $k$-dimensional space $K$, and
$$
\sum_{a,b}T_{ba}^\dagger T_{ba}=I_Q.
$$
The <operation fidelity> uses the unsquared <quantum fidelity> on a <purification of a density operator>, so its square is <entanglement fidelity>. The <Kraus formula for entanglement fidelity> gives
$$
F_{\rm op}(\mathcal D\mathcal C,\rho_Q)^2
=\sum_{a,b}|\operatorname{Tr}(\rho_Q T_{ba})|^2.
$$
Indeed each overlap $\langle\Psi_\rho|(I_R\otimes T_{ba})|\Psi_\rho\rangle$ equals $\operatorname{Tr}(\rho_QT_{ba})$.
For any one of these <Kraus operators> $T$, let $\Pi_T$ be the <orthogonal projection> onto its image, of <matrix rank> $r_T\leq k$. Since $\Pi_TT=T$, the <Cauchy-Schwarz inequality> for the <Hilbert-Schmidt inner product> gives the <rank bound for a weighted operator trace>:
$$
\begin{aligned}
|\operatorname{Tr}(\rho_QT)|^2
&=|\operatorname{Tr}[(\Pi_T\sqrt{\rho_Q})^\dagger(T\sqrt{\rho_Q})]|^2\\
&\leq\operatorname{Tr}(\rho_Q\Pi_T)\operatorname{Tr}(\rho_QT^\dagger T)\\
&\leq\left(\sum_{j<k}\lambda_j\right)\operatorname{Tr}(\rho_QT^\dagger T).
\end{aligned}
$$
The last step uses part (i) at $r_T$, followed by nonnegativity of the <eigenvalues> of the <density operator> $\rho_Q$. Summing over the <Kraus operators> and using their completeness relation proves \b[the <finite-dimensional quantum compression converse>]:
$$
\boxed{1-\epsilon\leq F_{\rm op}(\mathcal D\mathcal C,\rho_Q)^2\leq\sum_{j<k}\lambda_j.}
$$
No invertibility of $\rho_Q$, or restriction to an isometric decoder, was used.
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