Of course, the $\Leftarrow$ part of this is automatic, so the real condition is the $\Rightarrow$ part. In more elementary terms:

If $c \leq \bigvee D$ for $D$ a directed subset, then $c \leq d$ for some $d \in D$.

In the case where $P$ has a top element$1$, we say that $P$ is compact if $1$ is a compact element.

Examples

Given a set$X$, the finite elements of its power set are precisely the (Kuratowski)-finitesubsets of $X$. (This is the origin of the term ‘finite element’.)

If $R$ is a (not necessarily commutative) ring, the lattice $Idl(R)$ of two-sided ideals of $R$ is compact. Indeed, the top element is the ideal generated by $1$, the multiplicative identity, and $1 \in \bigvee_i I_i$ implies $1 \in I_i$ for some index $i$.