= Solution
The <comoving observer> measures $E=ap^0$, so $q=aE=a^2p^0$. Write $\mathcal H=a'/a$. The <null geodesic> condition implies
$$
(\delta_{ij}+h_{ij})p^ip^j=(p^0)^2.
$$
The time component of the <geodesic equation>, using $d\tau/d\lambda=p^0$, is therefore
$$
\frac{dp^0}{d\tau}
=-\mathcal Hp^0-\frac{\mathcal H(\delta_{ij}+h_{ij})p^ip^j+\tfrac12h'_{ij}p^ip^j}{p^0}
=-2\mathcal Hp^0-\frac{h'_{ij}p^ip^j}{2p^0}.
$$
Differentiate $q=a^2p^0$. The homogeneous expansion terms cancel:
$$
q'=2\mathcal Ha^2p^0+a^2\frac{dp^0}{d\tau}
=-\frac{a^2}{2p^0}h'_{ij}p^ip^j.
$$
At zeroth order $p^i/p^0=n^i$, with $\delta_{ij}n^in^j=1$. The difference between coordinate <velocity> and physical unit direction is already first order and multiplies $h'$, so it can be discarded in this expression. We obtain the <synchronous photon momentum redshift>
$$
\boxed{\frac{dq}{d\tau}=-\frac12q h'_{ij}n^in^j.}
$$
In the unperturbed universe $q$ is constant and physical <photon> energy redshifts as $a^{-1}$; the displayed term is the additional anisotropic first-order redshift.
To check the direction, let $v^i=p^i/p^0$. Combining the spatial and temporal <geodesic equations> cancels their homogeneous Hubble terms and gives
$$
\frac{dv^i}{d\tau}
=-h'^i{}_jv^j-\Gamma^i{}_{jk}v^jv^k
+\frac12v^i h'_{jk}v^jv^k=O(h).
$$
Every remaining connection or explicit metric-derivative term is first order. A local orthonormal spatial frame defines the physical unit vector by $n^i=v^i+\tfrac12h^i{}_jv^j+O(h^2)$. Its derivative adds only first-order terms, so
$$
\boxed{\frac{dn^i}{d\tau}=O(h).}
$$
Consequently direction deflection multiplied by an already first-order <temperature> anisotropy is second order. The first-order brightness calculation may use a straight unperturbed ray without neglecting the first-order gravitational frequency change.
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