= Solution
For an <adiabatic mode> the perturbation of baryon number per photon is zero. Since $n_\gamma\propto T^3$, $\rho_\gamma\propto T^4$ and nonrelativistic $\rho_b\propto n_b$, the common temperature/entropy perturbation gives $\delta_b=3\delta T/T$ and $\delta_\gamma=4\delta T/T$. Hence $\delta_\gamma=4\delta_b/3$.
Let $q=a n_e\sigma_T$, $\mathcal H=a'/a$ and $R=4\bar\rho_\gamma/(3\bar\rho_b)$. In <tight coupling>, the relative velocity is small but its drag force need not vanish, so it must be eliminated before setting the two velocities equal. Multiply the photon Euler equation by $R$ and add the baryon Euler equation. Their opposite drag terms cancel, even with the common drag-prefactor typo in the printed equations. The result at common velocity potential $\theta$ is
$$
(1+R)\theta'+\mathcal H\theta+R(\delta_\gamma/4-\sigma_\gamma)=0.
$$
Together with continuity this gives
$$
\boxed{\delta_\gamma'=\frac43k^2\theta,\qquad
\theta'=-3c_s^2\left(\frac14\delta_\gamma-\sigma_\gamma+\frac{\mathcal H}{R}\theta\right),}
$$
$$
\boxed{c_s^2=\frac{R}{3(1+R)}=\frac13\left(1+\frac{3\bar\rho_b}{4\bar\rho_\gamma}\right)^{-1}.}
$$
The same <photon-baryon sound speed> follows from $\delta P=\delta\rho_\gamma/3$ and $\delta\rho=\delta\rho_\gamma+\delta\rho_b$: using the adiabatic density relation gives $\delta P/\delta\rho=c_s^2$.
The Euler bracket in the original PDF prints $a/a'$; the derivation requires $a'/a=\mathcal H$. Its baryon continuity equation also needs a prime on $\delta_b$. Finally, for the defined velocity potential $\mathbf v=i\mathbf k\theta$, physical Thomson drag is $-q(\theta_\gamma-\theta_b)$, without an extra $k^{-2}$: taking a velocity divergence introduces $-k^2$ on both sides, which cancels on returning to the potential equation. This is <photon opacity normalization for a velocity potential>. These errors should not be concealed by reproducing a dimensionally inconsistent equation.
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