I was confused the existence of the intermediate von Neumann subalgebra $P$ in the above Lemma.
We know that $M^{\varphi}$ and $M$ must be globally invariant under $\sigma^{\varphi}$. Can we construct any another $P$ such that $P\neq M$ and $P\neq M^{\varphi}$, in addition, $P$ is globally invariant under $\sigma^{\varphi}$?

I don't have a concrete example, but at least I can give some heuristics why this is not unexpected at all.
Given $x\in M\setminus M^\varphi$ selfadjoint, you can construct (writing $\sigma$ instead of $\sigma^\varphi$ for clarity) $$ P=(M^\varphi\cup\{\sigma^\varphi_t(x):\ t\in\mathbb R\})''. $$ I guess without more concrete information about $x$ there might be situations where $P=M$, but at least it shows you a possibility for algebra that is invariant for $\sigma$ but different from $M^\varphi$.