= Solution
Before the gates, write the two <Bell states> as
$$
|\Phi^+\rangle_{12}|\Phi^+\rangle_{34}
=\frac12\sum_{a,b=0}^1|a\rangle_1|a\rangle_2|b\rangle_3|b\rangle_4.
$$
The <CNOT gate> sends the last bit to $b\oplus a$. Projection onto outcome 0 on <qubit> 4 therefore requires $b=a$. After the <Hadamard gate> on <qubit> 2, projection onto its outcome 0 contributes $1/\sqrt2$ for either remaining value of $a$. The unnormalized surviving state is
$$
\frac{|00\rangle_{13}+|11\rangle_{13}}{2\sqrt2}=\frac12|\Phi^+\rangle_{13}.
$$
Thus
$$
\boxed{|\psi\rangle_{13}=|\Phi^+\rangle_{13},\qquad P(00_{24})=\frac14.}
$$
This is <entanglement swapping>: <photons> 1 and 3 become entangled even though the joint measurement acts only on their original partners.
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