 Solution

With the convention that $p\le q$ means that $p$ is stronger, a set $G\subseteq \mathbb{P}$ is a generic filter over $M$ when it is a filter and meets every dense set $D\subseteq \mathbb{P}$ with $D\in M$. Explicitly, $G$ is upward closed toward weaker conditions, every two members have a common stronger member in $G$, and $G\cap D\ne \varnothing$ for every such $D$.