Let me take the following as a definition of weakly compact cardinal and show its equivalence to the embedding property:
Definition. $\kappa$ is weakly compact if for any collection $X$ of $\kappa$-many subsets of $\kappa,$ there exists a non-principal $\kappa$-complete filter on $\kappa$ deciding every element of $X$.
Theorem. The following are equivalent:
(1) $\kappa$ is weakly compact.
(2) $\kappa$ has the embedding property described in the question.
Proof. $(1) \to (2).$ Take a $\kappa$-model $M$. Let $X=P(\kappa) \cap M.$ Let $F$ be the filter produced by $(1)$ which decides every element of $X$. Take $M^\kappa/F.$ Note that this is well-founded (as $M$ is closed under $\omega$-sequences and $F$ is countably complete), so let $N$ be its transitive collapse and consider the resulting $j: M \to N.$ This witnesses $(2)$.
$(2) \to (1)$. Assume $X$ is given. Take $\kappa$-model $M$ with $X \in M, X \subset M.$ Let $j: M \to N$ witness $(2).$ Define $F$ by $F= \{ A \subset \kappa: A \in M, \kappa \in j(A) \}$. Then $F$ witnesses $(1)$ with respect to $X$.