The structure $L_{\omega_1^{CK}}$ consists of only HYP sets (I believe) and HYP in this structure is the same as the actual hyperaritmetic sets. Now if I move to the structure $L_{\omega_1^{CK}}[a]$ where $a$ is some non-hyperarithmetic real I add my guess is that HYP remains unaltered simply because HYP has a bottom up definition and won't be affected by the addition of new sets as long as one doesn't move to a non-$\omega$ model or the like.
Is this correct? Any more formal demonstration?
The following is not the answer of your question. But I think it is what you really want.
I guess you might be figuring out Leo's proof of McLaughlin's conjecture and his answer to Question 65 in Harvey's problem collection paper. The point is that by applying a nonstandard ordinal, Leo obtained a nonstandard $\Pi^0_1$-singleton so that it is not hyperarithmetic.
There are several ways to see above. One is by applying Barwise compactness. Another is to use Gandy's basis. By either way, you may obtain a nonstandard $\omega$-model $M$ of KP with $\omega_1^M=\omega_1^{CK}$ in which there is a nonhyperarithmetic (in the real sense) $\Pi^0_1$-singleton $x$ which has the property below.
To see it by Gandy's basis theorem. Just apply it to obtain an $\omega$-model $M\models KP$ in which $\omega_1^{CK}$ is nonstandard. In $M$, fix a nonstandard recursive ordinal $\alpha$, we can perform Leo's proof to produce a nonhyperarithmetic $\Pi^0_1$-singleton $x$so that for any $\beta<\alpha$, those reals computed by $x^{\beta}$ and $\emptyset^{\alpha}$ are precisely those computed by $\emptyset^{\beta}$.
Now take $N=L_{\omega_1^{CK}}[x]$. Since $x\leq_h M$, we have that $N\models KP$. It is not difficult to see that $N\models$ ''$x$ is a $\Pi^0_1$-singleton$"$.
The left is to show $N\models$`` $x$ is not hyperarithmetic".
$\bf{Proof}$: Otherwise, there must be some nonstandard ordinal $\gamma_0$ in $N$ so that $N\models \emptyset^{\gamma_0} $ exists. We may assume that $\gamma_0<\alpha$. Since $\omega_1^x=\omega_1^{CK}$, there must be some standard recursive ordinal $\gamma_1<\gamma_0$ so that $x^{\gamma_1}\geq_T \emptyset^{\gamma_0}$. Via some absoluteness (see below), it is not difficult to see $\emptyset^{\gamma_0}<_T \emptyset^{\alpha}$. Then by the property of $x$, $\emptyset^{\gamma_0}<\emptyset^{\gamma_1}$, a contradiction.[]
To see that $\emptyset^{\gamma_0}<_T \emptyset^{\alpha}$. There is recursive tree $T$ so that $N\models \emptyset^{\gamma_0} \mbox{ is the unique path in } T$. Note that there is also a real $z$, which is $\emptyset^{\gamma_0}$ in $M$, so that $M\models z \mbox{ is the unique path in } T$. If $\emptyset^{\gamma_0}=z$, then we have the conclusion. Otherwise, $\emptyset^{\gamma_0}$ does not belong to $M$. But $\emptyset^{\gamma_0}$ is hyperarithemtic in $x$ and so must belong to $M$, a contradiction.
In fact, by the proof above, every real which is hyperarithmetic in $N$ is actually hyperarithmetic.
