= Solution
Use <rovibrational spectroscopy> combination differences so that the band origin cancels. With the assignments above, $R(2)$ and $P(4)$ end at the same upper level $J'=3$, giving
$$
\widetilde\nu[R(2)]-\widetilde\nu[P(4)]
=B''\{4\cdot5-2\cdot3\}=14B''.
$$
Likewise $R(3)$ and $P(3)$ start at the same lower level $J=3$, so
$$
\widetilde\nu[R(3)]-\widetilde\nu[P(3)]
=B'\{4\cdot5-2\cdot3\}=14B'.
$$
Only the printed wavenumbers are needed:
$$
\boxed{B''=\frac{720.91-700.22}{14}=1.47786\ \mathrm{cm}^{-1},}
$$
$$
\boxed{B'=\frac{723.90-703.15}{14}=1.48214\ \mathrm{cm}^{-1}.}
$$
Thus \b[$B'-B''\simeq0.00429\ \mathrm{cm}^{-1}$]. The <rotational constants> differ only slightly, so the average <moment of inertia> changes little on vibrational excitation. The upper value is slightly larger, corresponding to a slightly smaller effective <moment of inertia>. This sign is possible for a bending excitation; the intuition that excitation must always lengthen a diatomic bond is not a general rule for a polyatomic bend.
The rounded printed values are internally consistent: fitting all six positions to the same elementary model gives approximately $\widetilde\nu_*=711.9945\ \mathrm{cm}^{-1}$, $B''=1.47797\ \mathrm{cm}^{-1}$ and $B'=1.48203\ \mathrm{cm}^{-1}$, with residuals below $0.002\ \mathrm{cm}^{-1}$. The displayed combination-difference values have more digits than their physical accuracy; the meaningful scale is about $1.48\ \mathrm{cm}^{-1}$.
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