= Solution
Use <energy> units $m=m_pc^2$, $M=m_Ac^2$, $e=E_\gamma$ and $P=pc=\sqrt{E^2-m^2}$. The <soft-photon resonant capture threshold> follows from <four-momentum conservation>:
$$
M^2=m^2+2e(E-P\cos\theta),\qquad
E-P\cos\theta=2E_0,\quad E_0=\frac{M^2-m^2}{4e}.
$$
For fixed <proton> <energy> the left side is largest at head-on incidence. Its threshold is $E+P=2E_0$. Combining that with $(E-P)(E+P)=m^2$ gives
$$
\boxed{E_{\min}=E_0+\frac{m^2}{4E_0},\qquad
\frac{E_{\min}}m=\frac{E_0}m+\frac m{4E_0}.}
$$
Equality is allowed for a head-on <photon>. Here $e=k_BT=2.3482\times10^{-4}\,\mathrm{eV}$, so
$$
\boxed{E_0=6.786\times10^{20}\,\mathrm{eV},\qquad E_{\min}/m\simeq7.23\times10^{11}.}
$$
A very soft <photon> needs an enormous <proton> <Lorentz factor> to reach the specified <invariant mass>. A true thermal <photon> distribution and a finite-width <resonance> would smear this monoenergetic threshold.
For $\theta=0$, $E-P=m^2/(E+P)\le m$, whereas $2E_0\gg m$, so production is impossible. For each nonzero angle, sufficiently large <energy> is possible, and $P\simeq E$ gives
$$
\boxed{E\simeq\frac{2E_0}{1-\cos\theta}.}
$$
To find the <maximum deflection in soft-photon capture>, note that the final particle's transverse <momentum> comes entirely from the <photon>:
$$
\tan\phi=\frac{e\sin\theta}{P+e\cos\theta}
\simeq\frac e{2E_0}\sin\theta(1-\cos\theta).
$$
Writing $x=\cos\theta$, maximization gives $(1-x)(2x+1)=0$ and the interior maximum $x=-1/2$. Therefore, to the same extraordinarily accurate ultrarelativistic order,
$$
\boxed{\theta_{\max}\simeq120^\circ,\qquad
\phi_{\max}\simeq\frac{3\sqrt3\,e}{8E_0}=2.25\times10^{-25}\,\mathrm{rad}.}
$$
The neglected relative corrections are of order $m^2/E_0^2$ and $e/E_0$, both around $10^{-24}$. The final particle is effectively collinear with the <proton> despite a large <photon> incidence angle.
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