Given a quantum system $|\psi\rangle=\alpha_0|\psi_0\rangle\otimes |0\rangle+\alpha_1|\psi_1\rangle\otimes |1\rangle$, such that each subsystem $|\psi_i\rangle$ is entangled with a qubit is state $|i\rangle$.
Is it correct that applying the unambiguous quantum state discrimination operator (USD) $P=I\otimes|0\rangle\langle0|$ on $|\psi\rangle$, where $I$ is the identity operator gives,
$P|\psi\rangle=|\psi_0\rangle\otimes |0\rangle$
We have $$P\left|\psi\right\rangle =I\otimes\left|0\right\rangle\left\langle0\right|\left(\alpha_0|\psi_0\rangle\otimes |0\rangle+\alpha_1|\psi_1\rangle\otimes |1\rangle\right)$$ $$=\alpha_0|\psi_0\rangle\otimes\left|0\right\rangle\left\langle0\right|0\rangle +\alpha_1|\psi_1\rangle\otimes\left|0\right\rangle\left\langle0\right|1\rangle =\alpha_0|\psi_0\rangle\otimes\left|0\right\rangle$$ if the states $\left|0\right\rangle$ and $\left|1\right\rangle$ are orthogonal (and then $\left\langle0\right|1\rangle=0$).