= Solution
Use the <two-pair component of polarization-entangled down-conversion> to account for identical <bosons> in the same collected modes. Write $a_H^\dagger$ for a <creation operator> in mode $a$ with <horizontal polarization>, and similarly for the other modes. For the plus <Bell state>, the pair-creation operator is $K^\dagger=a_H^\dagger b_V^\dagger+a_V^\dagger b_H^\dagger$. The normalized four-photon state is
$$
|\Psi_2\rangle=\frac{(K^\dagger)^2}{2\sqrt3}|0\rangle
=\frac{|2H_a,2V_b\rangle+|H_aV_a,H_bV_b\rangle+|2V_a,2H_b\rangle}{\sqrt3}.
$$
These are normalized <Fock states>. In particular, $(a_H^\dagger)^2|0\rangle=\sqrt2|2H_a\rangle$; omitting this factor would misnormalize the four-photon input.
The first <polarizing beam splitter> sends $a_H^\dagger$ to $T_H^\dagger$ and $a_V^\dagger$ to $c_V^\dagger$. Only the middle term can give exactly one trigger photon and leave another photon in $c$. In the first term both $a$ photons reach $T$ and none reaches $c$; in the last there is no trigger. Thus the relevant, still unnormalized component is
$$
\frac1{\sqrt3}T_H^\dagger c_V^\dagger b_H^\dagger b_V^\dagger|0\rangle.
$$
For the selected paths of the <half-wave plate> and <beam splitter>, take
$$
c_V^\dagger\longmapsto\frac{c_H^{\prime\dagger}+c_V^{\prime\dagger}}{\sqrt2},\qquad
b_\sigma^\dagger\longmapsto\frac{d_\sigma^\dagger+e_\sigma^\dagger}{\sqrt2},\quad\sigma=H,V.
$$
Here $e$ labels the output towards $D_3$. An immaterial common reflection phase has been absorbed in the mode definitions. Keeping one photon at $D_3$ leaves
$$
\frac{T_H^\dagger}{2\sqrt6}
(c_H^{\prime\dagger}+c_V^{\prime\dagger})
(d_H^\dagger e_V^\dagger+d_V^\dagger e_H^\dagger)|0\rangle.
$$
At the final <polarizing beam splitter>, the $H_{c'}H_d$ alternative yields one $H$ at each of $D_1,D_2$, while $V_{c'}V_d$ yields one $V$ at each. The two opposite-polarization alternatives send both photons to the same detector and fail fourfold <coincidence counting>. The accepted component is therefore
$$
\frac1{2\sqrt6}|H\rangle_T
\bigl(|HHV\rangle_{123}+|VVH\rangle_{123}\bigr).
$$
After conditioning and <normalization>, the three detected output modes have
$$
\boxed{|\mathrm{GHZ}\rangle_{123}=\frac{|VVH\rangle+|HHV\rangle}{\sqrt2}.}
$$
The <optical mode matching> must erase any timing, frequency or other record that distinguishes the two alternatives. Otherwise the same polarization populations could describe an incoherent <mixed quantum state>, rather than this <quantum superposition>. The <Bouwmeester three-photon GHZ experiment> uses this principle. The <photonic GHZ preparation by fourfold coincidence> is a measurement in the coincidence basis: detecting all four photons projects and consumes them. A click at $T$ by itself does not herald a guaranteed freely available three-photon state.
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