No generic is definable in a perfect notion of forcing of a model of Peano Arithmetic

Question

I would like to prove Lemma 6.1.2.2 from The Structure of Models of Peano Arithmetic by Kossak and Schmerl.

Let $\mathcal{M}$ be a countable model of Peano Arithmetic and $\mathbb{P}=\langle P, \le \rangle$ be a perfect ($\forall p \in P. \exists r_1, r_2 \in P. p \gt r_1, p \gt r_2$ and $r_1, r_2$ incompatible) notion of forcing with $P$ definable in $\mathcal{M}$. Then no generic is definable.

Answer 1

If a generic set $G\subset P$ were definable, then so would be its complement $P-G$. But this is a dense set, since every condition has incompatible extensions, and not both of them can be in $G$. So we'd have a definable dense set having no condition in $G$, which contradicts genericity.