Find the local unitary that takes the bell state to a state phi that has an extractable bell state

Question

I have a state $|p\rangle$ that has an extractable Bell state and I want to write it as a Bell state, $|b\rangle$, with a local unitary acting on one side. Basically I am trying to find a local unitary $U$ that satisfies:

$$ |p\rangle = (U \otimes 1)|b\rangle $$

$|b\rangle$ is a Bell state $|p\rangle$ is a known state with an extractable Bell state

Does anyone know how to do this?

My initial guess was $U \otimes 1 = |p \rangle \langle b|$ but this isn't a unitary operator.

The known state $|p\rangle$ is in state vector form. I am using Python an NumPy for reference.

Answer 1

$$ |p \rangle = ((aI + bZ + cX + dZX) \otimes 1) | \Phi^+ \rangle = a | \Phi^+ \rangle + b | \Phi^- \rangle + c | \Psi^+\rangle + d | \Psi^- \rangle\\ (aI+bZ+cX+dZX)(\bar{a}I+\bar{b}Z+\bar{c}X+\bar{d}XZ) = 1\\ a = \langle \Phi^+ | p \rangle\\ b = \langle \Phi^- | p \rangle\\ c = \langle \Psi^+ | p \rangle\\ d = \langle \Psi^- | p \rangle\\ $$

I'll leave the rest to you.