The probability of observing an outcome corresponding to $M_j$ (a positive measurement operator), when the quantum process has transformed some input state $\rho_j$ is


The evolution of a generic quantum state ρ under the channel E is obtained through the Choi matrix as $\mathcal{E}(\rho)=tr_{\sigma}[(\rho^T\otimes I_{\tau})\Lambda]$, where $\Lambda$ is the Choi matrix.

How can I get $p_{ij}$ by substituting $\mathcal{E}(\rho)$, cause there is already a partical trace in $\mathcal{E}(\rho)$, which will be $p_{ij}=tr\{tr_{\sigma}[(\rho^T\otimes I_{\tau})\Lambda]M_j\}$, seems terrible.


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The Choi is related to the channel via $$\Lambda = \sum_{ij} (E_{ij}\otimes \mathcal E(E_{ij})) = (I\otimes \mathcal E)\mathbb{P}_+,$$ using the shorthand notation $E_{ij}\equiv |i\rangle\!\langle j|$ and $\mathbb{P}_+\equiv |+\rangle\!\langle +|$, $|+\rangle\equiv \sum_i|i,i\rangle$.

As you already pointed out, you recover the channel $\mathcal E$ from $\Lambda$ via $$\mathcal E(\rho) = \operatorname{tr}_1[(\rho^T\otimes I)\Lambda].$$ Using these to compute probabilities you get $$\operatorname{tr}(M\mathcal E(\rho)) = \operatorname{tr}[M\operatorname{tr}_1[(\rho^T\otimes I)\Lambda]] = \operatorname{tr}[(\rho^T\otimes M) \Lambda].$$ Intuitively, you see this because the second space of $\Lambda$ is not affected by multiplication by $\rho^T$ nor by the partial trace, therefore the outer multiplication by $M$ directly connected to $\Lambda$.

More formally, you can show this expanding the partial trace explicitly: $$\operatorname{tr}_1[(\rho^T\otimes I)\Lambda] = \sum_i (\langle i|\otimes I) [(\rho^T\otimes I)\Lambda](|i\rangle\otimes I) = \sum_{ijk\ell} \rho_{ji} \Lambda_{jk,i\ell} E_{k\ell},$$ and thus $$\operatorname{tr}(M\mathcal E(\rho)) = \sum_{\alpha\beta} \bar M_{\alpha\beta} \mathcal E(\rho)_{\alpha\beta} = \sum_{ij\alpha\beta} \bar M_{\alpha\beta} \rho_{ji} \Lambda_{j\alpha,i\beta} = \operatorname{tr}[(\rho^T\otimes M)\Lambda].$$


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