= Solution
Use the unnormalized <Choi matrix>
$$
J_\Phi=(\Phi_A\otimes\mathrm{id}_B)(|\widetilde\Psi_{AB}\rangle\langle\widetilde\Psi_{AB}|).
$$
It is positive because $\Phi_A$ is a <completely positive map>. Take its spectral decomposition, absorbing square roots of nonzero eigenvalues into the vectors:
$$
J_\Phi=\sum_{k=1}^r|v_k\rangle\langle v_k|,\qquad r=\operatorname{rank}J_\Phi\leq d^2.
$$
Expand $|v_k\rangle=\sum_{i,j}v_{k,ij}|i\rangle_A|j\rangle_B$ and define the <Kraus operator> $A_k$ by $(A_k)_{ij}=v_{k,ij}$. Equivalently,
$$
|v_k\rangle=(A_k\otimes I_B)|\widetilde\Psi_{AB}\rangle.
$$
The preceding <relative state> identity then gives $(I\otimes\langle\phi_B^*|)v_k=A_k\phi_A$. Inserting the spectral decomposition into the stated reconstruction identity yields
$$
\Phi_A(|\phi\rangle\langle\phi|)=\sum_kA_k|\phi\rangle\langle\phi|A_k^\dagger.
$$
The conjugated index vector is required here; it is present in the PDF but missing in the converted TeX version of this identity.
Rank-one projectors span all operators by polarization. Explicitly, putting $P(w)=|w\rangle\langle w|$ gives
$$
4|u\rangle\langle v|=P(u+v)-P(u-v)+iP(u+iv)-iP(u-iv).
$$
Thus linearity extends the equality to every $X\in\mathcal B(\mathcal H_A)$:
$$
\boxed{\Phi_A(X)=\sum_kA_kXA_k^\dagger.}
$$
Finally trace preservation and cyclicity of the trace give
$$
\operatorname{Tr}X=\operatorname{Tr}\Phi_A(X)=\operatorname{Tr}\left[X\sum_kA_k^\dagger A_k\right].
$$
For pure-projector inputs this says the Hermitian operator $\sum_kA_k^\dagger A_k-I$ has zero expectation on every vector, so it vanishes. Therefore
$$
\boxed{\sum_kA_k^\dagger A_k=I_A.}
$$
This is the <spectral Kraus decomposition>. Using a normalized <Choi state> $J_\Phi/d$ instead would require the extra factor $\sqrt d$ when reshaping its eigenvectors into Kraus operators.
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