How big is the least non-$\Sigma^1_1$-pointwise-definable ordinal? There's a large countable ordinal which has cropped up (as a lower bound!) in a computable structure theory problem I'm playing with. At present I don't really understand how big it is, and I'm curious where it fits in amongst better-understood ordinals. (I do have a kind of upper bound, but it's weird and not very helpful to me.)
Say that $\alpha$ is $\Sigma^1_1$-pd iff for every $x\in L_\alpha$ there is some $\Sigma^1_1$ formula $\varphi$ in the language of set theory without parameters such that $\{x\}=\varphi^{L_\alpha}$. (Here "$\Sigma^1_1$" refers to quantification over subsets of $L_\alpha$, not $\omega$, so this is quite broad.) I'm interested in the following:

How large is the smallest non-$\Sigma^1_1$-pd  ordinal, $\eta$?

I'm interested in either  "theory-oriented" (e.g. "$\eta\ge$ the least $\alpha$ such that $L_\alpha\models\mathsf{KP\omega}$ + '$\omega_1$ exists'") or "stability-oriented" (e.g. "$\eta\le$ the least $\alpha$ such that $L_\alpha\prec_1L_{\omega_1}$") information about $\eta$. A genuine characterization would be great, but I suspect there isn't a snappy one; I'll settle for any interesting bounds (lower or upper) on $\eta$.
(To be fair there are a couple immediate projectum-flavored observations, namely that every $\Sigma^1_1$-pd ordinal is $\Pi^1_2$-projectible to $\omega$ and that if $\alpha$ is $\Sigma^1_1$-projectible to $\omega$ then $\alpha$ is $\Sigma^1_1$-pd, but that line of thought doesn't seem to give me much detailed information about $\eta$. So I'd be happy with projectum-flavored information as well, but I suspect that's the wrong way to go.)
 A: Claim: Let $\kappa$ be least such that $L_\kappa$ is admissible and
$L_\kappa\models$``$\omega_1$ exists''
and let $\alpha=\omega_1^{L_\kappa}$. Then $\alpha$ is the least
non-$\Sigma^1_1$-pd ordinal. Moreover, the 1st projectum of $L_\kappa$ is
$\rho_1^{L_\kappa}=\omega_1^{L_\kappa}=\alpha$, and the  1st standard parameter is $p_1^{L_\kappa}=\{\alpha\}$.
(Thus, $\kappa$ is also the least admissible such that
$\rho_1^{L_\kappa}>\omega$.)
The following lemma is probably standard, but I had not noticed it before:
Lemma: Let $\kappa$ be such that $L_\kappa$ is admissible and $\delta<\kappa$
such that $L_\kappa\models$``$\delta$ is a cardinal''. Then
$\rho_1^{L_\kappa}\geq\delta$.
Proof: Suppose not. Then $L_\kappa=\mathrm{Hull}_1^{L_\kappa}(\rho\cup\{p\})$
for some $\rho<\delta$ and $p\in L_\kappa$. (That is, the $\Sigma_1$-hull of
parameters in $\rho\cup\{p\}$.) Therefore $L_\kappa\models$``For every
$\beta<\delta$ there is an ordinal $\gamma$ such that
$\beta\in\mathrm{Hull}_1^{L_\gamma}(\rho\cup\{p\})$''. But then by
admissibility,
there is $\gamma<\kappa$ which works simultaneously for all $\beta<\delta$,
which yields a surjection $f:\rho\to\delta$ with $f\in L_\kappa$, contradicting
that $\delta$ was a cardinal there.
Proof of Claim: For the moment let $\alpha$ be any ordinal, and let
$\kappa_\alpha$ be the least $\kappa>\alpha$ such that $L_\kappa$ is admissible.
Let $\kappa=\kappa_\alpha$. Note that if there is
$\beta\in[\alpha,\kappa)$ such that $L_\beta$ projects to $\omega$, then one
can
identify the least such $\beta$ in a $\Sigma^1_1$-over-$L_\alpha$ way (using
the
usual relationship between illfounded models and admissibles), and therefore
$\alpha$ is $\Sigma^1_1$-pd.
So suppose that $\alpha=\omega_1^{L_\kappa}$, where $\kappa=\kappa_\alpha$.
Then
$L_\alpha\preceq_1 L_\kappa$, by condensation, and hence
$\mathrm{Hull}_1^{L_\kappa}(\alpha)=L_\alpha$. But
$L_\kappa=\mathrm{Hull}_1^{L_\kappa}(\alpha+1)$. For if this hull had ordinal
height
$\beta<\kappa$, just let $\psi$ be some $\Sigma_0$ formula and $p,d\in L_\beta$ such that
$L_\beta\models\forall x\in d\ \exists y\ \psi(p,x,y)$ but $\beta$ is least
such; then because $p,d$ are in the hull, so is $\beta$ actually,
contradiction. Combined with the lemma, this gives $\rho_1^{L_\kappa}=\alpha$
and
$p_1^{L_\kappa}=\{\alpha\}$.
Now suppose for a contradiction that $\alpha$ is $\Sigma^1_1$-pd. I claim
that $\rho_1^{L_\kappa}=\omega$ and $p_1^{L_\kappa}=\{\alpha\}$, contradicting
the previous paragraph. For this, we show that
$L_\kappa=\mathrm{Hull}_1^{L_\kappa}(\{\alpha\})$.
We can convert $\Sigma^1_1$-over-$L_\alpha$ to
$\Pi_1^{L_\kappa}(\{\alpha\})$ formulas (i.e. $\Pi_1$ over $L_\kappa$, in
parameter $\alpha$). (This must be a standard generalization of the situation when $\alpha=\omega$: Given a $\Sigma^1_1$ formula
$\varphi(x)=\exists y\psi(x)$, consider the game where player I plays elements
of $L_\alpha$ and player II also plays such elements, and player II also decides
which of these elements go into a predicate $y$, and then player II wins iff
the
elements played produce a structure $(M,\in,y)$ such that
$(M,\in,y)\models\psi(x)$.
Then for $x\in L_\alpha$, $(\varphi(x))^{L_\alpha}$ iff there is a winning
strategy for player II in the game iff the standard analysis of the game (see below) does not
yield
a winning strategy for player I in $L_\kappa$, and the latter is a
$\Pi_1^{L_\kappa}(\{\alpha\})$ statement.)
(Edit:) The analysis of the game: The game, at least once one sets up the rules appropriately, has open payoff for player I. Let $E$ be the set of partial plays of the game of even length. Let $S$ be the set of all $p\in E$ from which player I has a winning strategy (for the game continuing from $p$). We rank $S$, writing $S_{\beta}$ for the set of partial plays of rank ${<\beta}$. Let $S_0=\emptyset$ and $S_1$ be the set of $p\in E$ where player I has already won. For limit $\beta$ let $S_\beta=\bigcup_{\gamma<\beta}S_\gamma$. And for $\beta\geq 1$ let $S_{\beta+1}$ be the set of $p\in E$ such that there is $x\in L_\alpha$ such that for all $y\in L_\alpha$, we have $(p,x,y)\in S_\beta$. Note that $\beta\leq\gamma\implies S_\beta\subseteq S_\gamma\subseteq S$. Let $\beta_\infty$ be the least $\beta$ such that $S_\beta=S_{\beta+1}$. Then $S=S_{\beta_\infty}$. Now (by setting up the game rules appropriately) $E$ is definable over $L_\alpha$, and note that it follows that $S_\beta\in L_\kappa$ for each $\beta<\kappa$, and that $\left<S_\beta\right>_{\beta<\kappa}$ is $\Sigma_1^{L_\kappa}(\{\alpha\})$. Now $\beta_\infty\leq\kappa$. For otherwise we have some $p\in S_{\kappa+1}\backslash S_\kappa$, so for all $x\in L_\alpha$ there is $y\in L_\alpha$ such that $(p,x,y)\in\bigcup_{\beta<\kappa}S_\beta$, and note this contradicts admissibility. So player I wins iff $\emptyset\in S_\kappa$, and player II wins iff $\emptyset\notin S_\kappa$.
So, because $\alpha$ is
$\Sigma^1_1$-pd, for every $x\in L_\alpha$ there is a $\Pi_1$ formula $\varphi$
such that $x$ is the unique $x'\in L_\alpha$
such that $L_\kappa\models\varphi(x',\alpha)$. But then
$x\in\mathrm{Hull}_1^{L_\kappa}(\{\alpha\})$. For we have that for every
$x'\in L_\alpha$ with $x'\neq x$, $L_\kappa\models\neg\varphi(x',\alpha)$.
So by admissibility, there is $\beta<\kappa$ such that for every $x'\in 
L_\alpha$ with $x'\neq x$, we have $L_\beta\models\neg\varphi(x',\alpha)$. But then
note that the least such $\beta$ can be detected in a
$\Sigma_1^{L_\kappa}(\{\alpha\})$
manner, and from that we can compute $x$, so $x$ is in the hull, as desired.
This is a contradiction, showing that $\alpha$ is not $\Sigma^1_1$-pd.
The full Claim follows by putting the things above together above.
Remark: Let $\alpha$ be an ordinal and $\kappa=\kappa_\alpha$. Then conversely
to one point above, every $\Pi_1^{L_\kappa}(\{\alpha\})$ subset of $L_\kappa$ is
$\Sigma^1_1$-over-$L_\alpha$.
