Does the axiom of choice holds in $L(\mathcal{P}(\lambda))$? Cantor's Attic claimed the following:

However, Shelah proved that if λ is a strong limit cardinal of uncountable cofinality then $L(\mathcal{P}(\lambda))$ is a model of AC.

If that's because there is a definable global well-order of $L(\mathcal{P}(\lambda))$, it seems to imply that there is a global well-order of $V$, defined by $x \prec y$ if there is a strong limit cardinal of uncountable cofinality $\lambda$ such that $x \in L(\mathcal{P}(\lambda))$ and $y \notin L(\mathcal{P}(\lambda))$ or otherwise $x$ precedes $y$ in the $L(\mathcal{P}(\lambda))$-order, where $\lambda$ is the least strong limit cardinal of uncountable cofinality such that $x \in L(\mathcal{P}(\lambda))$.
This is too good to be true. What am I missing?
 A: You are missing the fact that the well-ordering of $\cal P(\lambda)$ is definable from a parameter, but indeed, many different parameters can be used. So in order to choose one for each $\lambda$ of uncountable cofinality, you have to use Global Choice to begin with.
Shelah's theorem really shows that in the presence of $\sf DC$, if $\lambda$ such that $\omega<\operatorname{cf}(\lambda)<\lambda$, then there is some $x$ such that $\cal P(\lambda)\subseteq\mathrm{HOD}_x$. This means that from this one $x$, we get a well-ordering of $\cal P(\lambda)$, and therefore a definable well-ordering of $L(\mathcal P(\lambda))$ using $x$ as a parameter. But, as I said, this $x$ is not quite definable in the general case. So in order to get $\sf AC$ you'd need, beyond $\sf DC$, also that you can choose such $x$ uniformly, which is begging the question.
Two relevant papers:

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*Shelah, Saharon, Set theory without choice: Not everything on cofinality is possible, Arch. Math. Logic 36, No. 2, 81-125 (1997). ZBL0877.03023.


*Cummings, James; Friedman, Sy-David; Magidor, Menachem; Rinot, Assaf; Sinapova, Dima, Ordinal definable subsets of singular cardinals, Isr. J. Math. 226, No. 2, 781-804 (2018). ZBL1436.03259.
