What is the cofinality of this ordinal (say in ZFC)? Is it countable or not?
Edit: Based on reason for close, I have added some informal motivation for question. Based on the computability-theoretic definition, whenever $\omega^{CK}_\alpha$ is countable then so is $\omega^{CK}_{\alpha+1}$. But, formally speaking, it doesn't apply directly when $\alpha$ isn't countable.
Denoting $\beta=\omega^{CK}_{\omega_1+1}$, intuitively there are several "interesting" and large enough points $p$ (such that $p<\beta$) so that $p$ will have countable finality (as one example when $p$ is the first fixed point of a normal function such as $x \mapsto (\omega_1)^x$).
But this informal intuition doesn't help in giving a formal answer to cofinality of $\omega^{CK}_{\omega_1+1}$. Hence the reason for asking the question (I want to know how well the informal intuition does or doesn't transfer into formal reasoning).
I disagree with Stefan Mesken's answer. First, let me explain why I think the general claim made there is false. Consider the theory $T$ = KP + V=L + "There is some ordinal $\alpha$ such that $\alpha+\alpha$ doesn't exist but $\alpha+\beta$ exists for all $\beta<\alpha$." We have $L_{\omega_1+\omega_1}\models T$, but no $\alpha\in (\omega_1,\omega_1+\omega_1)$ models $T$; so $L_{\omega_1+\omega_1}$ is the first level of $L$ satisfying $T$ above $\omega_1$, yet $cf(\omega_1+\omega_1)\not=\omega$.
Now here's an argument showing that the cofinality of $\omega_{\omega_1+1}^{CK}$ - which latter ordinal I'll call "$\theta$" for simplicity - is $\omega_1$.
The key is projecta. If $\alpha$ is the next admissible above $\beta$, then there is an injection $\pi$ of $\alpha$ into $\beta$ which is $\Sigma_1$-definable (in $L_\alpha$). This doesn't contradict admissibility of $\alpha$ since in order to turn this into a $\Sigma_1$ cofinal map from $\beta$ to $\alpha$ we would need $ran(\pi)$ to be $\Delta_1$, and it's only $\Sigma_1$.
Indeed, $ran(\pi)$ has the following strong property: for any cofinal set $A\subseteq \alpha$, $\pi[A]$ is not $\Delta_1$. Otherwise we could indeed get a $\Sigma_1$ map from $\beta$ to $\alpha$ with cofinal image - namely, send $x\in A$ to $\pi^{-1}(x)$ and $x\not\in A$ to $17$. This is in fact $\Sigma_1$, since if we know ahead of time that $x\in ran(\pi)$ we can search for $x$'s preimage without risk of looping forever.
OK, now suppose $cf(\theta)=\omega$. Fix$^1$ a constructible cofinal $\omega$-sequence $(\eta_i)_{\eta<\omega}$, and let $\gamma_i=\pi(\eta_i)$. Then $S=(\gamma_i)_{i\in\omega}$ is an $\omega$-sequence of countable ordinals, so by condensation$^1$ we get $S\in L_{\omega_1}$. But now take $S$ - or rather, the range of $S$ - to be our $A$.
(OK fine, strictly speaking all I've shown is that $cf(\theta)>\omega$. But since $\vert\theta\vert=\omega_1$, this does mean $cf(\theta)=\omega_1$.)
$^1$Why does this exist? On the face of things, after all, I seem to be using V=L (or at least that $L$ correctly computes cofinalities) here. This isn't necessary, but at present I'm forgetting how the argument goes; I'll add it when I have time to recall it.