Which consistency conditions in the degeneracy subspace $\mathcal{E}_i$ would you find if the perturbation $V$ does not remove the degeneracy in that subspace in first order?
Solution. The derivation of (9.21) shows that if we still have $E_{i \alpha}^{(1)}=E_{i \beta}^{(1)}$ for all degeneracy indices in $\mathcal{E}_i$, consistency of the second order equation requires that not just the operator $V_i^{(1)}=\mathcal{P}_i^{(0)} V \mathcal{P}_i^{(0)}$ is diagonal, but also that the operator
$$
V_i^{(2)}=\mathcal{P}_i^{(0)} V \frac{1-\mathcal{P}_i^{(0)}}{E_i^{(0)}-H_0} V \mathcal{P}_i^{(0)}
$$
is diagonal. However, consistency of the simultaneous diagonalization of $V_i^{(1)}$ and $V_i^{(2)}$ then also implies the condition
$$
\left[V_i^{(1)}, V_i^{(2)}\right]=\mathcal{P}_i^{(0)} V\left(\mathcal{P}_i^{(0)} V \frac{1-\mathcal{P}_i^{(0)}}{E_i^{(0)}-H_0}-\frac{1-\mathcal{P}_i^{(0)}}{E_i^{(0)}-H_0} V \mathcal{P}_i^{(0)}\right) V \mathcal{P}_i^{(0)}=0 .
$$
If $V$ preserves the degeneracy in $\mathcal{E}_i$, but these consistency conditions cannot be met, then $H=H_0+V$ apparently does not have a complete set of eigenstates which scale analytically under scaling $V \rightarrow \lambda V$ of the perturbation.