= Solution
Use the squared total <energy-momentum four-vector>, expressed in energy-squared units. With the <proton> initially at rest,
$$
W^2=(m_pc^2+E_\gamma)^2-E_\gamma^2=m_p^2c^4+2m_pc^2E_\gamma.
$$
At the <reaction threshold>, the outgoing <proton> and <pion> have no relative <kinetic energy> in their <centre-of-momentum frame>, so $W=(m_p+m_\pi)c^2$. Solving gives
$$
\boxed{E_\gamma^{\rm th}=m_\pi c^2\left(1+\frac{m_\pi}{2m_p}\right)=144.715\,\mathrm{MeV}.}
$$
The numerical value uses the supplied $938$ and $135\,\mathrm{MeV}$ <rest energies>. The threshold exceeds the <pion>'s <rest energy> because the outgoing system must carry the incident <photon>'s nonzero laboratory <momentum>.
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