= Solution
Write the total derivative of the <phase-space distribution function> as
$$
\frac{df}{d\eta}=\frac{\partial f}{\partial\eta}
+\frac{dx^i}{d\eta}\frac{\partial f}{\partial x^i}
+\frac{d\ln\epsilon}{d\eta}\frac{\partial f}{\partial\ln\epsilon}
+\frac{d\widehat p^i}{d\eta}\frac{\partial f}{\partial\widehat p^i}.
$$
Since $a\bar T$ is constant for the background photon gas, expand the <Bose-Einstein distribution> as
$$
f=\bar f(\epsilon)-\epsilon\bar f_{,\epsilon}\Theta+O(2).
$$
At first order the four terms are respectively
$$
-\epsilon\bar f_{,\epsilon}\Theta',
\qquad
-\epsilon\bar f_{,\epsilon}\widehat p^i\partial_i\Theta,
\qquad
\epsilon\bar f_{,\epsilon}
(\Phi'-\widehat p^i\partial_i\Psi),
\qquad
0.
$$
The angular-deflection velocity is already first order and multiplies the first-order angular dependence of $f$, so its contribution is second order. Thus the collisionless left-hand side of the <Free-streaming photon Boltzmann equation> is
$$
\boxed{\frac{df}{d\eta}
=-\epsilon\bar f_{,\epsilon}
\left[\Theta'+\widehat{\mathbf p}\cdot\nabla\Theta
+\widehat{\mathbf p}\cdot\nabla\Psi-\Phi'\right]}.
$$
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