= Solution
In <Newtonian gauge>, the coordinate-comoving observer has normalized four-velocity $u^\mu=a^{-1}(1-\psi,\mathbf0)$, hence $u_0=a(1+\psi)$. In the $(+---)$ convention the measured photon energy is $E=u_\mu p^\mu$. Thus \b[to first order]
$$
\boxed{E=a(1+\psi)\frac\epsilon{a^2}(1-\psi)=\frac\epsilon a+O(\psi^2).}
$$
Use the time component of the <geodesic equation> and divide by $p^0=d\eta/d\lambda$:
$$
\frac{dp^0}{d\eta}=-\Gamma^0{}_{\alpha\beta}\frac{p^\alpha p^\beta}{p^0}.
$$
Let $\mathcal H=a'/a$ and take the photon path direction $\mathbf e$ at zeroth order. The left side divided by $\epsilon/a^2$ is $\epsilon'/\epsilon-2\mathcal H+2\mathcal H\psi-d\psi/d\eta$. The contraction on the right side, using the provided connection coefficients, is
$$
-2\mathcal H+2\mathcal H\psi-\partial_\eta\psi
-2\mathbf e\cdot\nabla\psi+\partial_\eta\phi.
$$
The homogeneous expansion cancels. Therefore \b[the scalar photon energy-redshift law is]
$$
\boxed{\frac{d\ln\epsilon}{d\eta}=\partial_\eta\phi-\mathbf e\cdot\nabla\psi
=-\frac{d\psi}{d\eta}+\partial_\eta\phi+\partial_\eta\psi.}
$$
The total derivative is $d/d\eta=\partial_\eta+\mathbf e\cdot\nabla$ on the unperturbed ray. Perturbing that path inside a first-order potential would contribute only at second order. The homogeneous $a^{-1}$ energy redshift is already removed by the comoving momentum variable $\epsilon$.
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