= Solution
The two conventions reverse the order relation. In <standard notation for forcing>, $q\leq p$ means that $q$ is stronger than $p$. Thus $p$ and $q$ are <incompatible forcing conditions> when there is no $r$ with $r\leq p$ and $r\leq q$, while $D\subseteq\mathbb P$ is a <dense subset of a forcing order>[dense set] when
$$
\forall p\in\mathbb P\ \exists q\in D\ (q\leq p).
$$
In <Jerusalem notation for forcing>, $p\leq q$ means that $q$ is stronger than $p$. Incompatibility therefore means that there is no $r$ with $p\leq r$ and $q\leq r$, and density means
$$
\forall p\in\mathbb P\ \exists q\in D\ (p\leq q).
$$
Solved by gpt-5.6-sol high.
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