The following example with a little modified from the handbook of set theoretic topology, Page 574:
Let $\kappa$ be any cardinal for which there exists a family $\{H_\alpha: \alpha < \kappa\}$ of infinite subset of $\omega$ such that
1) $\beta < \alpha$ implies $|H_\beta \setminus H_\alpha|<\omega$ and
2) for every $H \in [\omega]^\omega$, there exists $\alpha <\kappa$ such that $ |H\setminus H_\alpha|=\omega$.
The underlying set for the desired space $X$ is the set $\kappa \cup \omega$, where we consider $\kappa$ and $\omega$ to be disjoint. The topology is definied as follows: For each $\alpha < \kappa$ we define a basic neighbourhood of $\alpha$ for each $\beta < \alpha$ and each $F\in [\omega]^{<\omega}$ by $$ N(\alpha: \beta, F)=\{ \text{ any set } A \subset \kappa \text{ with } \alpha \in A\} \cup ((H_\beta \setminus H_\alpha)\setminus F).$$ Points in $\omega$ are declared to be isolated. With this topology on $X=\kappa \cup \omega$ it is clear that $\omega$ is a countable dense set of isolated points, and $\kappa$ is a closed discrete subset of $X$ whose subspace topology is the usual discrete topology on $\kappa$.
Question: Is $X$ pseudocompact?
Thanks for your help.