You have a small typo in (2): it should say "below some $p\in G$" rather than $p\in P$. And I guess that you should also say in (2) that $G$ is nonempty.
If $G$ satisfies (1), then it satisfies (2) because if $p$ is in $G$ and $D$ is dense below $p$, then let $D'$ be the set of things that conditions $q$ which are either in $D$, D$or incompatible with$p$. This is dense in$P$. P$ since any condition that is compatible with $p$ will have elements of $D$ below it, and any condition incompatible with $p$ is already in $D'$. But $G$ cannot meet $D'$ in something incompatible with $p$, by your assumption on $G$, and so it must meet it in $D$. D$, as desired. Conversely, if$G$is nonempty and satisfies (2), then it will satisfy (1) because if$D$is dense, then it is dense below any$p$, and so$G$will meet it. 1 You have a small typo in (2): it should say "below some$p\in G$" rather than$p\in P$. And I guess that you should also say in (2) that$G$is nonempty. If$G$satisfies (1), then it satisfies (2) because if$p$is in$G$and$D$is dense below$p$, then let$D'$be the set of things that are either in$D$, or incompatible with$p$. This is dense in$P$. But$G$cannot meet$D'$in something incompatible with$p$, so it must meet it in$D$. Conversely, if$G$is nonempty and satisfies (2), then it will satisfy (1) because if$D$is dense, then it is dense below any$p$, and so$G\$ will meet it.