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There's even a more direct way than the one described by Adam. Note that every pure Choi state corresponds to a superoperator acting by conjugation. Since $$ \mathbb{I} \otimes \mathbb{I} |\phi^+\rangle = |\phi^+\rangle, \qquad Z \otimes \mathbb{I} |\phi^+\rangle = |\phi^-\rangle, $$ the inverse of $c_\lambda$ under the Choi-Jamiołkowski isomorphism is simply $$ T_\lambda = \lambda\, \mathrm{id} + (1-\lambda)\, Z \cdot Z^\dagger. $$ This gives you the Kraus operators $$ K_1 := \sqrt{\lambda}\,\mathbb{I}, \qquad K_2:= \sqrt{1-\lambda}\, Z. $$