Expanding the normalized <maximally entangled state> in the definition of the <Choi state> gives
$$
C_{\widetilde\Lambda}
=\frac1d\sum_{k,\ell}\widetilde\Lambda(|k\rangle\langle\ell|)\otimes|k\rangle\langle\ell|
=\boxed{\sum_jp_j|a_j\rangle\langle a_j|\otimes|\overline{b_j}\rangle\langle\overline{b_j}|}.
$$
Here the bar denotes componentwise <complex conjugation> in the basis defining the <Choi state>. Every term is a positive <tensor-product operator>, and the whole operator has trace one because $\widetilde\Lambda$ is assumed to be a <quantum channel>. It is therefore a <separable quantum state>, proving entanglement breaking by the <separable Choi-state criterion for entanglement breaking>.

If the vectors are unit vectors, the displayed $p_j$ are already the product-state weights. If they are not normalized, absorb their squared norms into the weights and normalize the nonzero vectors. Trace preservation requires
$$
d\sum_jp_j\lVert a_j\rVert^2|b_j\rangle\langle b_j|=I;
$$
the probability-distribution condition alone would not guarantee this. Equivalently, the <Kraus operators> are $K_j=\sqrt{dp_j}|a_j\rangle\langle b_j|$, exhibiting the <rank-one Kraus representation of an entanglement-breaking channel>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 323 / 1 / vi / Solution

An <entanglement-breaking channel> admits a <Kraus representation> with rank-one <Kraus operators> $K_j=|a_j\rangle\langle b_j|$. Its <Choi state> is a sum of positive <tensor-product operators>, hence separable. Conversely, spectral decompositions of each prepared state and each <POVM> effect in a <measure-and-prepare channel> produce rank-one Kraus operators. Trace preservation imposes $\sum_j\lVert a_j\rVert^2|b_j\rangle\langle b_j|=I$.

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 Incoming links: Rank-one Kraus representation of an entanglement-breaking channel