I just wanted to fix my answer, which I couldn't do yesterday as it was already midnight and I was too tired (nevertheless the answer already given by Amit is elegant and true)
As $D$ is normal $\kappa$ is represented in $M \cong Ult_{D} (V)$ by the diagonal function $ d: \kappa \to \kappa$, and as $\kappa$ is measurable, we only need $d$ to be defined as the identity on the $\aleph_{\alpha}$'s. Hence the cardinal $2^{\kappa}$ is represented in each element of $M$ is already determined by the partial a function defined only on the cardinals below kappa.
Now if $\kappa$, x \in P(\kappa)^{M}$ then there exists a function $g: h: \aleph_{\alpha} kappa \to 2^{\aleph_{\alpha}}$, . Then with the hint we know V$ such that $ f_{D} x = h_{D}$, and as $M \aleph_{\kappa + j(\beta)models h_{D} = \aleph_{\kappa + subset \beta}$ in $M$, and by our assumption kappa$ it follows that {$ \gamma $\aleph_{\alpha} < \kappa \, : h (\aleph_{\alpha}) \, f(\gamma) = g(\gamma)$} subset \aleph_{\alpha}$} $\in D$. Thus $M \models P(\kappa) \subset g_{D}$ where $g_D$ denotes the equivalence class of the function $g: \aleph_{\alpha} \to P(\aleph_{\alpha})$. This means that leads us to $M$ thinks that M \models |P(\kappa)| \le |g_{D}|$. But the elements cardinal $|g_{D}|$ is represented by $f$ and $g$ are the same, hence function $M f: \models aleph_{\alpha} \to 2^{\aleph_{\alpha}}$.
Invoking the hint we may conclude$$M\models 2^{\kappa} \le f_{D} \le \aleph_{\kappa + \beta}$$and as $P(\kappa)^{M} = P(\kappa)$ we finally have $2^{\kappa} \le (2^{\kappa})^{M} \le (\aleph_{\kappa + \beta})^{M} \le \aleph_{\kappa + \beta}$.
I hope this is helpfulbeta}$
