= Solution
The comoving observer's <four-velocity> has $u_0=-a$, so its measured photon energy is $E=-u_\mu p^\mu=ap^0$, and the comoving momentum is $q=aE=a^2p^0$. Let $P^i=p^i/p^0=dx^i/d\tau$. The <null geodesic> condition is
$$
(\delta_{ij}+h_{ij})p^ip^j=(p^0)^2.
$$
Divide the temporal <geodesic equation> by $p^0=d\tau/d\lambda$, and use the supplied <Christoffel symbols>:
$$
\frac{dp^0}{d\tau}
=-\mathcal H p^0-\mathcal H\frac{(\delta_{ij}+h_{ij})p^ip^j}{p^0}
-\frac12h'_{ij}\frac{p^ip^j}{p^0}
=-2\mathcal H p^0-\frac12p^0h'_{ij}P^iP^j.
$$
Differentiating $q=a^2p^0$ cancels the homogeneous redshift terms. In the remaining first-order term, replace $P^i$ by the unperturbed unit propagation direction $n^i$:
$$
\boxed{\frac{dq}{d\tau}=-\frac12q h'_{ij}n^in^j+O(h^2).}
$$
This is the <synchronous photon momentum redshift>. Background physical momentum decreases as $a^{-1}$, while the background comoving momentum stays constant.
The spatial <geodesic equation> gives $dp^i/d\tau=-2\mathcal H p^i$ in the unperturbed universe. Since the temporal equation gives the identical factor for $p^0$, their ratio $P^i$ is constant at zeroth order. More explicitly, at first order,
$$
\frac{dP^i}{d\tau}=-h'^i{}_j n^j-\Gamma^i{}_{jk}n^jn^k
+\frac12n^i h'_{jk}n^jn^k+O(h^2).
$$
Every displayed term contains a <metric perturbation> or its derivative. The orthonormal-frame unit direction differs from $P^i$ by a first-order metric correction, $n^i=P^i+h^i{}_jP^j/2+O(h^2)$, whose derivative is also first order. Hence \b[$dn^i/d\tau=O(h)$], with no background change of direction. Direction corrections multiplied by an already first-order source contribute only at second order.
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