= Solution
The reference, singly excited and doubly excited determinants are mutually orthogonal. Singles have amplitude $O(\lambda)$ and doubles $O(\lambda^2)$, so only the reference norm and squares of single-excitation amplitudes contribute through second order:
$$
\langle\Phi^{(2)}|\Phi^{(2)}\rangle
=1+\sum_{ia}(\lambda_i^a)^2+O(\lambda^4).
$$
There is no reference-double cross term. The stated expansion has no cubic contribution because its single and double sectors are orthogonal.
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