= Solution
{c}
Use the ordered polarization <basis> $(|H\rangle,|V\rangle)$ and define the <Bell states>
$$
|\Phi^\pm\rangle=\frac{|HH\rangle\pm|VV\rangle}{\sqrt2},\qquad
|\Psi^\pm\rangle=\frac{|HV\rangle\pm|VH\rangle}{\sqrt2}.
$$
Regrouping the four-<photon> <quantum state> by channel pairs $(1,4)$ and $(2,3)$ gives
$$
\begin{aligned}
|\Phi^+\rangle_{12}|\Phi^+\rangle_{34}=\frac12\big(&|\Phi^+\rangle_{14}|\Phi^+\rangle_{23}
+|\Phi^-\rangle_{14}|\Phi^-\rangle_{23}\\
&+|\Psi^+\rangle_{14}|\Psi^+\rangle_{23}
+|\Psi^-\rangle_{14}|\Psi^-\rangle_{23}\big).
\end{aligned}
$$
For instance, projecting channels 2 and 3 on $\langle\Psi^-|$ retains the two original terms with inner polarizations $HV$ and $VH$, with opposite signs, leaving $(|HV\rangle-|VH\rangle)_{14}/(2\sqrt2)$. Its squared norm is $1/4$.
By the same <orthonormality> argument, each <Bell-basis measurement> outcome prepares the identically named <Bell state> in channels 1 and 4, with
$$
\boxed{P(\Phi^+)=P(\Phi^-)=P(\Psi^+)=P(\Psi^-)=\frac14.}
$$
Before communication of the outcome, Alice and Bob's <density matrix> is $I_4/4$; the usable conditional <entanglement> is established by the announced <Bell-basis measurement> result. This assumes an ideal complete <Bell-basis measurement>; ordinary passive linear optics alone does not deterministically distinguish all four polarization <Bell states>.
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