ZF without regularity allows P(A) member of A Last week, I was doing a bit of reading around on some axiom of regularity(/foundation)-related questions, and found an answer in one of them (which I cannot seem to locate, for the life of me, right now) which stated that, although it was a fairly technical result, one could show, in ZF-reg, that $P(A) \in A$ is possible.
Well, that sort of made my brain fizz a little, and so I've been thinking about it for the past few days, but cannot find any documentation relating to the result. If anyone could send me in the right direction, or provide an example or outline of such, it would be much appreciated!

No, it most certainly does not imply that $\mathcal P(A)\in A$. Simply because $\mathcal P(\varnothing)\notin\varnothing$.
Moreover, you get an entire model of $\sf ZF$ inside any model of $\sf ZF-Reg$. So there will be plenty of sets which do not have $\mathcal P(A)$ as their element.
But it is consistent. The easiest way would be to take one of the "Anti Foundations Axioms" which assert that the failure of regularity (known as foundations as well) is extensive. How extensive? For example, "Every possible graph is realized". Then you only need to show that you can draw a graph of a set and its power set, such that the power set is an element of the set.
You can do it directly using permutation models. I cannot give a reference, since I learned it in class a few years ago, and the notes are in Hebrew, but one of the exercises was to show that it is consistent to have $\mathcal P(A)\in A$. If my memory serves me right, $A$ had three elements.
