A sequence of ultrafilters of $\omega$ (therefore each $p_n \in \beta \omega$) $p_n$ is discrete if there exists a sequence $X_n$ of subsets of $\omega$ such that for each $n$, $X_n \in p_n$ and if $n \neq m$ then $X_n\cap X_m=\emptyset$.
Let $p_n$ be a discrete sequence of ultrafilters on $\omega$. Let $q \in \text{cl} \{p_n: n \in \omega\}$. Then we define $\Omega(p, q)=\{A\subset \omega: \forall B \in q \exists n \in A(B \in p_n)\}$.
I must show that $r=\Omega(p, q)$ is an ultrafilter. Clearly, $\emptyset \notin r$ and $r$ is upward closed and since $q \in \text{cl} \{p_n: n \in \omega\}$ it follows that $\omega \in \Omega(p, q)$
More details: Given $A \subset \omega$
It remains to see that $r$ is closed under finite intersections and that the filter is maximal. Can someone help?