The importance of antichains in forcing is that for most purposes, dense sets and maximal antichains are equivalent. A maximal antichain $A$ is one that cannot be extended and still be an antichain. This means every element of $p ∈ P$ is compatible with some member of $A$. Their existence follows from Zorn's lemma. Given a maximal antichain $A$, let $D = \{ p : ( \exists q \in A ) ( p≤q ) \}$. $D$ is dense, and $G∩D≠0$ if and only if $G∩A≠0$. Conversely, given a dense set $D$, Zorn's lemma shows there exists a maximal antichain $A⊆D$, and then $G∩D≠0$ if and only if $G∩A≠0$.
https://en.wikipedia.org/wiki/Forcing_(mathematics)#The_countable_chain_condition
I really don't understand what this is saying. First of all,
This means every element of p ∈ P is compatible with some member of A.
I mean, how can "every" $p$ is compatible with some member of $A$ when $A$ is antichain? Aren't elements of $A$ also in $P$? Then how can members of $A$ be compatible with each other when the definition of antichain says they are not compatible?
Since $A$ is a maximal antichain, then for every $p \in P$ there are two options: either $p \in A$ (whence $p$ is compatible with an element of $A$, namely itself), or $p \notin A$ (whence $A \cup \{ p \}$ is not an antichain, meaning that there are distinct $q,r \in A \cup \{ p \}$ which are compatible, and since $A$ is an antichain exactly one of $q$ and $r$ is $p$).
The definition of an antichain is that any two dinstinct elements of that set are incompatible; every condition is clearly compatible with itself.