= Solution
For <resonance channel threshold conversion>, equate the compound-state <energy> measured from the two different channel thresholds. The proton-plus-nitrogen threshold exceeds the alpha-plus-carbon threshold by
$$
\Delta=(938.26+13973.30-3728.34-11177.76)\,\mathrm{MeV}=5.46\,\mathrm{MeV}.
$$
Thus $T_p^*=T_\alpha^*-\Delta=2.113770\,\mathrm{MeV}$ and
$$
\boxed{T_{p,\mathrm{lab}}=T_p^*\frac{938.26+13973.30}{13973.30}=2.25570\,\mathrm{MeV}\simeq2.26\,\mathrm{MeV}}.
$$
The positive centre-of-momentum <energy> verifies that the <proton> channel is open. These numbers use the stipulated nonrelativistic kinematics and rounded nuclear masses.
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