In "A Trichotomy Theorem in Natural Models of $AD^+$" by Caicedo and Ketchersid, they write as a consequence of the Moschovakis Coding Lemma, "This yield that if $M$ and $N$ are transitive models of $AD$ with the same reals, and $\gamma < \min\{\Theta^M, \Theta^N\}$, then $\mathcal{P}(\gamma)^M = \mathcal{P}(\gamma)^N$."
Here $\Theta$ is sup of the length of prewellorderings on $\mathbb{R}$. The question is how to see this remark.
Let $\leq$ be a prewellordering of length $\gamma$. Let $\mathbf{\Sigma}_1^1(\leq)$ denote the smallest pointclass closed under $\exists^\omega$, $\forall^\omega$, $\wedge$, $\vee$, continuous reductions, and $\exists^\mathbb{R}$. This point class has a universal set $U \subseteq \mathbb{R} \times \mathbb{R}$. A consequence of the Moschovakis coding lemma says that for all $S \subseteq \gamma$, there is some $x \in \mathbb{R}$ so that $\xi \in S$ if only if $(\exists y)(|y|_{\leq} = \xi \wedge U(x,y))$.
Now suppose $M$ and $N$ are two transitive models of $AD$ with the same reals. Let $\gamma < \min\{\Theta^M,\Theta^N\}$. If there exists a PWO $\leq$ of length $\gamma$ with $\leq \in M$ and $\leq \in N$, then one can show $\mathbf{\Sigma}^1_1(\leq) \subseteq M \cap N$. In particular the universal set $U$ is in both $M$ and $N$. If $S \subseteq \gamma$ and $S \in M$, by the coding lemma, there is some $x \in \mathbb{R}$ so that $U_x$ codes $S$ in the sense above. Since $N$ has $U$ and all the reals of $M$ (in particular $x$), $U_x$ codes $S$ in $N$. So $S \in N$.
The argument above seem to require that there is a common PWO $\leq$ of length $\gamma$ in both models. Is this generally true? Or is there some other way that the Moschovakis coding lemma is used to prove this result?
Yes, there seems to be a hypothesis missing here. By a result of Woodin, if there exists a Woodin cardinal which is a limit of Woodin cardinals, then, in a forcing extension there exist models $M$ and $N$ of $\mathsf{AD}^{+}$ with the same reals, each with a set of reals not in the other. This appears as Theorem 6.1 in Ilijas Farah's article in volume 4 of the LNL's reissue of the Cabal Seminar volumes.
Given a cardinal $\lambda$, a set of reals is $\lambda$-Suslin if it is the projection of a tree on $\omega \times \lambda$. The cardinal $\lambda$ is a Suslin cardinal if some set of reals is $\lambda$-Suslin but not $\gamma$-Suslin for any $\gamma < \lambda$.
Let $\Gamma = \mathcal{P}(\mathbb{R})^{M} \cap \mathcal{P}(\mathbb{R})^{N}$. By another theorem of Woodin, reproved by Trevor Wilson, the model $L(\Gamma)$ satisfies the statement that the Suslin cardinals are cofinal in the $\Theta$ of $L(\Gamma)$ (call it $\kappa$, and note that $\kappa$ is less than both $\Theta^{M}$ and $\Theta^{N}$). I don't know if a proof of this result appears in the literature. The Suslin cardinals being cofinal implies that every set of reals is Suslin (a proof of this appears in the version of "Extensions of the Axiom of Determinacy" on my webpage). This implies that for every set of reals $A$ in $L(\Gamma)$, $A$ is $\lambda$-Suslin in $L(\Gamma)$ for some $\lambda < \kappa$.
By another theorem of Woodin (whose proof appears in the solo article by Ketchersid in the same volume where the Caicedo-Ketchersid article appears), $\mathsf{AD}^{+}$ implies that the Suslin cardinals are closed below $\Theta$. So, in both $M$ and $N$, $\kappa$ is a Suslin cardinal. There exist then trees on $\omega \times \kappa$ in these models witnessing this fact. These trees are constructible from subsets of $\kappa$, and their projections cannot be in $L(\Gamma)$. So each of of $M$ and $N$ contains a subset of $\kappa$ not in the other.