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Precise interpretability strength of $\mathcal P_{DF}(\omega)$ and $L_{\omega_1^{CK}}$

I am curious about the relationship between the definable power set of $\omega$ and the $\omega_1^{CK}$th level of the constructible sets $L$.

In short, $\omega_1^{CK}$ is the least nonrecursive ordinal (the least recursive ordinal with no computable description in Kleene's $\mathcal O$, a hyperarithmetical set), and $L$ is a model of $ZFC$ constructed by a weakening of the power set axiom in the following manner:

Let $\mathcal P_{DF}(S) := \{X_i \subseteq S: {X_i}$ is definable from first-order parameters in ${S}\}$. To construct $L$, we have:

$L_0$ := $\varnothing$

$L_{\kappa+1}$ := $\mathcal P_{DF}(L_\kappa)$

$L_\lambda$ := $\bigcup\limits_{\kappa < \lambda} L_\kappa$ ($\lambda$ = limit ordinal).

and $L$ := $\bigcup\limits_{\kappa \in ORD} L_\kappa$

It has been remarked that $\mathcal O$ is the existential theory of $L_{\omega_1^{CK}}$. Clearly, $\omega \models PA$, so it is natural that $\mathcal P_{DF}(\omega) \models PA$ since all the recursive sets are contained in the standard system of first-order definable sets, denoted $SSY(\omega)$, and thus would appear at some countable level in $L$ by definition. Furthermore, we can see that $\mathcal P_{DF}(\omega) \subseteq L_{\omega_1^{CK}}$ follows also by definition of $L$.

My question is this: Can we say that $\mathcal P_{DF}(\omega)$ has the same relative interpretability power as $L_{\omega_1^{CK}}$ by virtue of its first-order definable closure of the recursive sets, in the same way that $\mathcal O$ is the existential theory of $L_{\omega_1^{CK}}$ as it preserves all "$\mathbf \Sigma^{1}_{1}$ -canonical" properties of the recursive sets at the $\omega_1^{CK}$th level in $L$?

If not, it would appear that $L_{\omega_1^{CK}}$ is "just above" $\mathcal P_{DF}(\omega)$ in its interpretability strength, but I am unsure how to make this notion of being "just above" anymore precise without a vague reference to $\mathcal O$ being the existential theory of $L_{\omega_1^{CK}}$.

tww
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