As Noah wrote in the comments, Shelah really means that $N$ is in the domain of $\mathbb D$. However some extra motivation for what's going on here might make things clearer. The point of the definition is to generalize $\sigma$-closed forcing. Shelah wants to iterate forcing notions that don't add reals. On the one hand, $\sigma$-closed forcing is the class of forcing notions par excellence that does not add reals but on the other hand $\sigma$-closed forcing is so non-destructive that there are limitations to what can be proved using it (for instance, $\sigma$-closed forcing does not specialize Aronszajn trees, it does not add branches to $\omega_1$-trees etc).

The relevant property Shelah wants to generalize is that if $\mathbb P$ is $\sigma$-closed, $\mu$ is sufficiently large and $N \prec H_\mu$ is a countable model containing $\mathbb P$ and some condition $p$ then $any$ generic $p \in G \subseteq \mathbb P \cap N$ over $N$ has a lower bound (it can be argued that this is what's really used in a lot of arguments involving $\sigma$-closed forcing). Roughly the idea for the generalization is to replace $all$ generics by $most$ generics (say a "measure one set") i.e. an element of a suitable filter of subsets of $Gen(N, \mathbb P, p)$.

To make this precise, we need to make sense of what types of filters we're going to use and this is what $\mathbb D(N, \mathbb P, p)$ is giving us. Now, the point is that if $\mu$ is sufficiently large (say $(2^{|\mathbb P|})^+$) then whether or not a generic $G \subseteq \mathbb P \cap N$ has a lower bound for $N \prec H_{\mu^*}$ for $\mu^* > \mu$ really only depends on $N \cap H_\mu$. Moreover, if $\mu \in N$ then an easy Tarski-Vaught argument tells you that $N \cap H_\mu \prec H_\mu$ so the filter of generics we get from $\mathbb D (N, \mathbb P, p)$ really need not be any different from the filter of generics we get from $\mathbb D(N\cap H_\mu, \mathbb P, p)$.

Uri Abraham's chapter in the handbook also discusses $\mathbb D$-complete forcing and the definitions and discussion there might help clarify things as well.

the domain of$\mathbb{D}$ ... include[s] $N$:" the goal is for $\mathbb{D}(N,P,p)$ to be defined at all, even when $N$ might be too big for the original definition of $\mathbb{D}$. $\endgroup$ – Noah Schweber Dec 27 '20 at 21:52extendan initial function $\mathbb{D}'$, which in an abuse of notation is conflated with $\mathbb{D}$ itself to be defined on a larger domain. $\endgroup$ – Noah Schweber Dec 27 '20 at 22:24