= Solution
For <heralded Bell-pair projection with two polarization parity filters>, a <Hadamard gate> followed by <Computational basis>[computational-basis] detection is a measurement in the diagonal <polarization> basis $|\pm\rangle=(|H\rangle\pm|V\rangle)/\sqrt2$. It can therefore be realized directly by a <polarizing beam splitter> at $45^\circ$ with a <photodetector> in each output: its transmitted and reflected clicks label $+$ and $-$. No separate <Hadamard gate> plate is required for this measurement.
Conditioning on one <photon> in each output of the first <polarizing beam splitter> implements the even $Z$-parity filter
$$
\Pi_Z^{25}=|00\rangle\langle00|+|11\rangle\langle11|=\frac{I+Z_2Z_5}{2}.
$$
The $45^\circ$ <polarizing beam splitter> similarly implements the even $X$-parity filter $\Pi_X^{46}=(I+X_4X_6)/2$. These are ideal coherent filters, with the optical <phases> chosen consistently. The all-zero <photodetector> record projects the upper outputs onto $\langle++|_{25}$ and the lower outputs onto $\langle00|_{46}$. The two effective bras are
$$
\langle++|\Pi_Z=\frac{\langle00|+\langle11|}{2}=\frac{\langle\Phi^+|}{\sqrt2},\qquad
\langle00|\Pi_X=\frac{\langle00|+\langle11|}{2}=\frac{\langle\Phi^+|}{\sqrt2}.
$$
Therefore the unnormalized retained state is
$$
\begin{aligned}
|\chi\rangle_{13}
&=\frac12\langle\Phi^+|_{25}\langle\Phi^+|_{46}
\bigl(|\Phi^+\rangle_{12}|\Phi^+\rangle_{34}|\Phi^+\rangle_{56}\bigr)\\
&=\frac18|\Phi^+\rangle_{13}.
\end{aligned}
$$
For example, the simultaneous Bell contractions require the three pair bits to coincide, leaving only the $000$ and $111$ terms of the original pair expansion. Each surviving $|00\rangle_{13}$ or $|11\rangle_{13}$ amplitude is $1/(8\sqrt2)$. Consequently
$$
\boxed{|\psi\rangle_{13}=|\Phi^+\rangle_{13},\qquad
P(0000\text{ and valid output paths})=\frac1{64}.}
$$
This <probability> is conditioned on production of all three input pairs, but not yet on successful <polarizing beam splitter> routing. The routing <probability> itself is $\langle\Pi_Z^{25}\Pi_X^{46}\rangle=1/4$: each nonidentity expectation vanishes on the relevant initial reduced state. Thus, if the question instead regards the routing <postselection> as already performed,
$$
\boxed{P(0000\mid\text{valid output paths})=\frac{1/64}{1/4}=\frac1{16}.}
$$
These two denominators describe different conditioning. Neither includes the <probability> that the three crystals actually emit the required pairs, <photodetector> inefficiency or undesired multipair events.
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