Every $B \in V$ in some $L[A]$? There are two standard versions of relative constructibility - $L(A)$ and $L[A]$. For any set $B$ it is trivial to find an $A$ such that $B \in L(A)$ - just take $A=B$. But except in specific cases, for example if $B$ is a set of ordinals, we don't have $B \in L[B]$.
Is it possible to have $B$ such that for all $A\in V$ $B \notin L[A]$?
My thoughts so far - it is known that for all $A$ there exists $B\subset ORD$ such that $L[A]=L[B]$. So basically if we can have $L(\mathscr P(ORD))\neq V$, meaning that there is an inner model of $V$ that has all subsets of ordinals, but isn't equal to $V$ itself, then any $B\in V\setminus L(\mathscr P(ORD))$ would be an example for such a set.
Also, for such a $B$ to exist we must have $V\vDash \neg AC$, because if $V$ has choice, then there must be a set of ordinals in $V\setminus L(\mathscr P(ORD))$.
 A: You're right. If $B$ cannot be well-ordered, then $B$ cannot be an element of any model of the form $L[A]$, since those are all models of choice.
Of course, if $B$ cannot be well-ordered, then $\{B\}$ cannot be in $L[A]$, despite being well-orderable. I mean, for crying out loud, it's a singleton!
So what would be a reasonable limitation? The transitive closure, of course. If $\operatorname{tcl}(\{B\})$ can be well-ordered, then we can code it as a set of ordinals $A$ and then $B\in L[A]$ for obvious reasons. But if it cannot be well-ordered, then there is no way to find $B$ in a transitive model of $\sf AC$.
As for the model $L(\mathcal P(\mathrm{Ord}))$, recall the Balcar–Vopěnka theorem:

If $M$ and $N$ are two models of $\sf ZF$ with the same sets of ordinals, and one of them satisfies $\sf AC$, then $M=N$.

(Often times it is just assumed that both are models of $\sf ZFC$, but in fact you only need one of them to be a model of $\sf AC$.)
And indeed, it is consistent that $M$ and $N$ are both models of $\sf ZF+\lnot AC$, but they have the same sets of ordinals. One classical example is the Feferman model and Cohen's first model. In both cases you add $\omega$ Cohen reals over $L$, but whereas in the Feferman model you only add the reals (and things constructible from finitely many of them), in the Cohen model you also add the set itself. It is not hard to show that in both cases, every set of ordinals is definable from a real, and since these have the same reals, they have the same sets of ordinals.

Let me also point out that the theorem above can be extended to arbitrary iteration of the power set operation. Namely, we can talk about sets-of-sets-of-sets of ordinals, and so on, and we can formulate a correct principle which is a weakening of choice in that sense. You can find the details in my thesis papers, but more specifically in:

Asaf Karagila, The Bristol Model: an abyss called a Cohen real. arXiv:1704.06939

Specifically, in section 5. But the whole construction is a witnessing of the fact that without any sort of choice, no $\mathcal P^\alpha(\mathrm{Ord})$ can give you a characterization of the universe.
