$(\kappa, \kappa, 2)$-saturated ideals?

Suppose $\kappa$ is regular uncountable and $I$ is a $\kappa$-aditive ideal on $\kappa$ such that forcing with $I$ is $\kappa$-Knaster. Force with $I$: Let $G$ be a generic filter and $j:V \to M \subseteq V[G]$ be the generic embedding with critical point $\kappa$ ($M$ is the well founded generic ultrapower). Let $T$ be a normal $\kappa$-tree in $V$. Let $B \subseteq T$ be a cofinal branch in $M$ (look at $j(T)$). Choose $\{p_i : i < \kappa\}$ such that $p_i$ decides the $i$th level node of $B$ and use $\kappa$-Knaster to get a branch in $V$. So $\kappa$ has the tree property. Add inaccessibility and you get weak compactness.

This should be a comment:

"Is it consistent to have a $(\kappa,\kappa, 2)$-saturated ideal at an inaccessible at all?"

This is well-known to be equiconsistent with "$\exists $ measurable" and could happen at small large cardinals - E.g., there could be a cardinal $\kappa < \mathfrak{c}$ and a $\kappa$-additive ideal $I$ on $\kappa$ such that forcing with $I$ is sigma-centered. The tree property continues to hold as explained.

Sorry, I missed the inaccessible there. Assuming you want an atomless forcing here, I think this is impossible.

For let $I$ be an atomless $\kappa$-additive $\kappa$-Knaster ideal on $\kappa$. Contruct a tree $T = \langle t_{\sigma} : \sigma \in 2^{< \kappa}\rangle$ of $I$-positive sets as follows. At successor step, if $t_{\sigma}$ is $I$-positive, $t_{\sigma0}, t_{\sigma1}$ split it into $I$-positive sets. At limits take intersection. Some branches could die because of $I$-null intersection but inaccessibility of $\kappa$ lets the construction run for $\kappa$ levels. This gives us a $\kappa$-tree with no branch (otherwise $\kappa$-cc is violated). But this is impossible as we argued that $\kappa$ had the tree property.