Another question on saturated models of ZFC Let $M$ be a saturated model of ZFC and let $\kappa$ be any cardinal in $M$. Now let
$$\begin{align*}
p_\kappa(f) &= \{ “f \text{ 1-1 function”} \wedge  \operatorname{dom}(f) \subseteq \omega \wedge \operatorname{ran}(f) \subseteq \kappa \} \\ 
&\qquad\cup \{n \in \operatorname{dom}(f) : n \in \omega \} \cup \{ \alpha \in \operatorname{ran}(f) : \alpha \in \kappa \}.
\end{align*}$$
$p_\kappa(f)$ is fin. realized in $M$ and therefore it is realized in $M$ (since $M$ is saturated and $\kappa < \operatorname{card}(M)$).
So there exist $f_\kappa : \omega \to \kappa$ that is $1-1$ and onto, which of course is a contradiction. What am I doing wrong?
Note : This is question is related to another question that I asked earlier, "A question on saturated models of ZFC". I don't know if I should merge the two questions.
 A: The problem with the argument you propose is that your type has too many parameters.  When you say that $M$ is saturated, then you mean that it has some cardinality $\delta=|M|$, and any finitely consistent type using fewer than $\delta$ many parameters is realized in the model.  Your type uses as parameters all the nonstandard natural numbers $n$ of $\omega$, meaning the $\omega$ of $M$ and so this includes all the nonstandard natural numbers of $M$, as well as the ordinals $\alpha$ that $M$ thinks are less than $\kappa$, which you fixed.   
For your argument to work, you need that this amounts to fewer than $\delta$ many parameters.  But it is not.  In fact, if $M$ is saturated, then there is an easier type showing that $M$ must have $|M|$ many nonstandard natural numbers.  The reason is that otherwise we can write down the type asserting that $x$ is a natural number not among them, and this type will be finitely realized by not realized in $M$. 
Meanwhile, you type is indeed consistent, and it is realized in another model. In fact, it is realized in the forcing extension of $M$ collapsing $\kappa$ to become countable. 
