If $M$ is HOD of a model $N$ of $\text{AD}^+ + V=L(P(ℝ))$, what kind of forcing construction in $M$ gives back such an $N$?
HODs for $\text{AD}^+ + V=L(P(ℝ))$ are conjectured (and under anti-large cardinal assumptions, proven) to be fine-structural models, so there should be a canonical construction here (as opposed to there being little say besides $N$ being a symmetric generic extension of $M$). In the ordinary derived model construction, $ω_1^N$ would be a limit of Woodin cardinals of the base model, but here (to get $M=HOD^N$) we want some kind of alternative derived model construction making $ω_1^N$ the least measurable in $M$ instead.
Some background and conjectures: The axiom of determinacy is central to our understanding of natural sets of reals. $Θ$ is the supremum of the ordinals that are a surjective image of the reals. Below long extenders, for countable models, being HOD for a model of a fragment of third order arithmetic ($Σ^2_1$ comprehension, or full comprehension if the HOD satisfies the separation schema; $Θ = Ord$ here) + ADR is perhaps equivalent to being a self-iterable fine-structural model (specifically, least branch premouse, as defined by relevant first order properties) with a proper class of cutpoint Woodin cardinals (thus it cannot have a strong cardinal, but superstrong cardinals should be ok). Under $\text{AD}^+$, every set of reals is ∞-Borel, encoded by a bounded subset of $Θ$. Under $\text{AD}^+$, if LSA (Largest Suslin Axiom) fails or conjecturally if $Θ$ is not a Woodin limit of Woodin cardinals in $HOD^{L(P(ℝ))}$, then every set of reals is definable from a countable set of ordinals. LSA fails in the model consisting of all universally Baire sets of reals (assuming a proper class of Woodin cardinals), but it is important in the hierarchy of $\text{AD}^+$ models to get there.
For the question, it should suffice to give a forcing construction to add an appropriate bounded subset of $Θ$, and then iterate $ω$ times, adding appropriate $x_0, x_1, ...$ with sufficiently fast-growing $M$-degrees, and take the direct limit. "Sufficiently fast-growing" can be enforced through forcing conditions.
To add an appropriate set, one possibility is the following. I assume familiarity with the extender algebra (link). Let $δ$ be a cutpoint Woodin cardinal in $M$, and we want to add an $ω$ sequence of ordinals $S$ with $\sup(S) < δ$. Consider formulas of size $<δ$ in the infinitary propositional logic with ordinals $<δ$ as constants. Call $S$ $M$-consistent iff for each extender and the corresponding elementary embedding $j$, and infinitary formula in $M$ (as above) $φ$, $j(φ)(j(S))⇔ φ(S)$. Adding $M$-consistent $S$ to $M$ should be appropriate; $δ$ being Woodin should give genericity, while the relevant closure and self-iterability should allow $δ$ to stay Woodin and make "sufficiently fast-growing" work well.
If $Θ=θ_0$ (i.e. $N ⊨ V=HOD(ℝ)$), then $x_i$ can just be reals.
Possibilities for LSA: If $M$ satisfies LSA and $δ=Θ$ and $κ$ is the least $<δ$-strong, $ω$ sequences might not suffice. However, we can choose an elementary embedding $j:M→M'$ with $\operatorname{crit}(j)=κ$, and take $x_0⊂κ$ from $M'[S]$ for an $M'$-consistent $S$, and perhaps require $x_0$ to be witnessed this way by $α$-strong embeddings for arbitrarily high $α < δ$; I do not know whether that works. The problem with simply allowing $M$-uncountable $S$ is that we do not know what $j(S)$ should be (though having $S$ as disjoint union of $ω$ sets in $M$ would be ok). There is a version of the extender algebra construction that does not use $j(S)$ (in the link above), but we want to preserve even the first measurable cardinal.
