I wanted to see for which ranks Bernays' Reflection Principle holds; that is, for every class and every property (allowing quantification over all classes) which is true about that class, there is a transitive set $u$ for which $\mathcal{P}(u)$ satisfies that property about that class's intersection with $u$. Like Vopenka did with Vopenka cardinals, I allow "arbitrary $A\subseteq V_\kappa$" to be classes in $V_\kappa$.
It turns out that once analyzed, these ranks bear a striking similarity to those ranks of Indescribable cardinals. I formalized this definition in ZFC as follows:
Let $\kappa$ be $\Pi_m$-Bernays when for every first-order $\Pi_m$ sentence $\phi$ in the language $\{\in,P\}$ where $P$ is an unary predicate symbol:
$$\forall A\in V_{\kappa+1}(\langle V_{\kappa+1};\in,A\rangle\models\phi\rightarrow\exists u\in V_\kappa(u\;\mathrm{is}\;\mathrm{transitive}\land\langle \mathcal{P}(u);\in,A\cap u\rangle\models\phi))$$
A similar definition in ZFC would be that for every first-order $\Pi_m$ unary formula $\phi$ in the language $\mathcal{L}_\in$:
$$\forall A\subseteq V_\kappa(\phi^{V_{\kappa+1}}(A)\rightarrow\exists u\in V_\kappa(u\;\mathrm{is}\;\mathrm{transitive}\land\phi^{\mathcal{P}(u)}(A\cap u)))$$
Of course, the existence of a $\Pi_{<\omega}$-Bernays cardinal is equivalent to the existence of a cardinal rank which satisfies Bernays' Reflection Principle. Thus, $\Pi_{<\omega}$-Bernays cardinals, in consistency strength, are somewhere above Bernays' Reflection Principle, which oddly enough implies the existence of an Inaccessible.
These cardinals are therefore consistency-wise stronger than that of Inaccessible cardinals.
Every cardinal is $\Pi_0$-Bernays. This is pretty simple once one considers that all of the unary first-order formulas in the language $\mathcal{L}_\in$ are either true for every set or true for no set.
This brings me to my question:
Are these cardinals inconsistent with ZFC? Are they any weaker or stronger than Indescribability?