The question you've stated isn't the question in Jech, you've made a minor typo. Here's the actual problem:
If $\beta < \kappa$ and {$\aleph _{\alpha} : 2^{\aleph _{\alpha}} \leq \aleph _{\alpha + \beta}$} $\in D$ and $D$ is a normal measure on $\kappa$, then $2^{\aleph _{\kappa}} \leq \aleph _{\kappa + \beta}$
Note that since $\kappa$ is measurable, $\aleph _{\kappa} = \kappa$.Also, note that they don't really need to tell you explicitly that $\beta < \kappa$; given the rest of the assumptions $\beta < \kappa$ would follow from inaccessibility of $\kappa$.
Okay, now we know that a normal measure extends the club filter, and the set of cardinals below $\kappa$ is club in $\kappa$, hence it makes sense in the hint to define $f(\aleph _{\alpha}) = \aleph _{\alpha + \beta}$ without specifying how $f$ acts on non-cardinals. Following my comment, let $g(\aleph _{\alpha}) = 2^{\aleph _{\alpha}}$. Then $g \leq f$ almost everywhere, and so:
$M \vDash [g] \leq [f]$
i.e.
$M \vDash j(g)(\kappa) \leq j(f)(\kappa)$
i.e.
$M \vDash 2^{\kappa} \leq \aleph _{\kappa + j(\beta)}$
Since $\beta < \kappa$, $j(\beta) = \beta$. Thus there is an injection from $(2^{\kappa})^M$ to $\aleph _{\kappa + \beta} ^M$. Since $P(\kappa) = P^M(\kappa)$, it means there's an injection from $2^{\kappa}$ to $\aleph _{\kappa + \beta}^M$. Finally, $\aleph _{\kappa + \beta} ^M \leq \aleph _{\kappa + \beta}$ since $M \subseteq V$.
