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 $X$.

Let me directly show $\Pi^1_1$-indescribability implies the embedding property (following Hauser's proof). Thus let $\kappa$ be $\Pi^1_1$-indescirable but assume it does not satisfy the embedding property as witnessed by the $\kappa$-model $M$. Let $E \subset \kappa \times \kappa$ code $(M, \in)$ such that if $\pi$ is the transitive collapse map, then $\pi(0)=\kappa.$ Let $F=\pi \restriction \kappa$ and $T=\{(n, \bar{\xi}): n$ is a (code of a) first order formula that holds in $(\kappa, E)$ under the assignment $\bar{\xi} \}$.

Let $\Phi(\kappa, E, F, T)$ be the formula:

$\Phi(\kappa, E, F, T)$ $(\kappa, E)$ is well founded and extensional $F=\pi \restriction \kappa$, $\pi(0)=\kappa$ and $T$ is the first order theory of $(\kappa, E)$.

Consider the $\Pi^1_1$-sentence which is satisfied over $V_\kappa:$

$\forall M [M, \kappa$-model of size $\kappa$ and $\kappa, E, F ,T \in M \implies $ $M \models \Phi(\kappa, E, F, T)$ but there is no $N, j$ with $|N|=\kappa$ and $j$ from transitive collapse of $(\kappa, E)$ into $N$ with critical point $\kappa].$

By $\Pi^1_1$-indescribability, there is inaccessible $\lambda < \kappa$ such that $(\lambda, E \cap \lambda \times \lambda)$ is well founded and extensional, $F \cap \lambda \times \lambda=$ (transitive collapse map)$^{-1} \restriction \lambda$ which sends $0$ to $\lambda$, and $T \cap \omega \times \lambda^{<\omega}$ is the theory of $(\lambda, E \cap \lambda \times \lambda)$.

Let $(M^*, \in)$ be the transitive collapse of $(\lambda, E \cap \lambda \times \lambda)$. Then we have $j^*: M^* \to M$ with $crit(j^*)=\lambda$ and $j^*(\lambda)=\kappa.$ Let $X \prec M$ with $X$ a $\lambda$-model and $j^*[M^*] \cup \{ \lambda\} \subset X$. Let $k: X \to N$ be the transitive collapse map and $j=k \circ j^*: M^* \to N.$ But then $j$ and $N$ witness that the above sentence fails in $V_\lambda$ a contradiction.

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 $X$.

Let me directly show $\Pi^1_1$-indescribability implies the embedding property (following Hauser's proof). Thus let $\kappa$ be $\Pi^1_1$-indescribable but assume it does not satisfy the embedding property as witnessed by the $\kappa$-model $M$. Let $E \subset \kappa \times \kappa$ code $(M, \in)$ such that if $\pi$ is the transitive collapse map, then $\pi(0)=\kappa.$ Let $F=\pi \restriction \kappa$ and $T=\{(n, \bar{\xi}): n$ is a (code of a) first order formula that holds in $(\kappa, E)$ under the assignment $\bar{\xi} \}$.

Let $\Phi(\kappa, E, F, T)$ be the formula:

$\Phi(\kappa, E, F, T)$ $(\kappa, E)$ is well founded and extensional $F=\pi \restriction \kappa$, $\pi(0)=\kappa$ and $T$ is the first order theory of $(\kappa, E)$.

Consider the $\Pi^1_1$-sentence which is satisfied over $V_\kappa:$

$\forall M [M, \kappa$-model of size $\kappa$ and $\kappa, E, F ,T \in M \implies $ $M \models \Phi(\kappa, E, F, T)$ but there is no $N, j$ with $|N|=\kappa$ and $j$ from transitive collapse of $(\kappa, E)$ into $N$ with critical point $\kappa].$

By $\Pi^1_1$-indescribability, there is inaccessible $\lambda < \kappa$ such that $(\lambda, E \cap \lambda \times \lambda)$ is well founded and extensional, $F \cap \lambda \times \lambda=$ (transitive collapse map)$^{-1} \restriction \lambda$ which sends $0$ to $\lambda$, and $T \cap \omega \times \lambda^{<\omega}$ is the theory of $(\lambda, E \cap \lambda \times \lambda)$.

Let $(M^*, \in)$ be the transitive collapse of $(\lambda, E \cap \lambda \times \lambda)$. Then we have $j^*: M^* \to M$ with $crit(j^*)=\lambda$ and $j^*(\lambda)=\kappa.$ Let $X \prec M$ with $X$ a $\lambda$-model and $j^*[M^*] \cup \{ \lambda\} \subset X$. Let $k: X \to N$ be the transitive collapse map and $j=k \circ j^*: M^* \to N.$ But then $j$ and $N$ witness that the above sentence fails in $V_\lambda$ a contradiction.

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 $X$.

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 $X$.

Let me directly show $\Pi^1_1$-indescribability implies the embedding property (following Hauser's proof). Thus let $\kappa$ be $\Pi^1_1$-indescirable but assume it does not satisfy the embedding property as witnessed by the $\kappa$-model $M$. Let $E \subset \kappa \times \kappa$ code $(M, \in)$ such that if $\pi$ is the transitive collapse map, then $\pi(0)=\kappa.$ Let $F=\pi \restriction \kappa$ and $T=\{(n, \bar{\xi}): n$ is a (code of a) first order formula that holds in $(\kappa, E)$ under the assignment $\bar{\xi} \}$.

Let $\Phi(\kappa, E, F, T)$ be the formula:

$\Phi(\kappa, E, F, T)$ $(\kappa, E)$ is well founded and extensional $F=\pi \restriction \kappa$, $\pi(0)=\kappa$ and $T$ is the first order theory of $(\kappa, E)$.

Consider the $\Pi^1_1$-sentence which is satisfied over $V_\kappa:$

$\forall M [M, \kappa$-model of size $\kappa$ and $\kappa, E, F ,T \in M \implies $ $M \models \Phi(\kappa, E, F, T)$ but there is no $N, j$ with $|N|=\kappa$ and $j$ from transitive collapse of $(\kappa, E)$ into $N$ with critical point $\kappa].$

By $\Pi^1_1$-indescribability, there is inaccessible $\lambda < \kappa$ such that $(\lambda, E \cap \lambda \times \lambda)$ is well founded and extensional, $F \cap \lambda \times \lambda=$ (transitive collapse map)$^{-1} \restriction \lambda$ which sends $0$ to $\lambda$, and $T \cap \omega \times \lambda^{<\omega}$ is the theory of $(\lambda, E \cap \lambda \times \lambda)$.

Let $(M^*, \in)$ be the transitive collapse of $(\lambda, E \cap \lambda \times \lambda)$. Then we have $j^*: M^* \to M$ with $crit(j^*)=\lambda$ and $j^*(\lambda)=\kappa.$ Let $X \prec M$ with $X$ a $\lambda$-model and $j^*[M^*] \cup \{ \lambda\} \subset X$. Let $k: X \to N$ be the transitive collapse map and $j=k \circ j^*: M^* \to N.$ But then $j$ and $N$ witness that the above sentence fails in $V_\lambda$ a contradiction.