I thought you will also need no cardinal between P(P(A)) and P(P(P(A))) because, I believe, latter is the upper bound on Hartogs' number of A. Also, thanks for link to the paper!

In your paper, you put the question "does $\Sigma=\Sigma_C$?". This is false, because $\Sigma$ is computable from $0^\blacktriangledown$, as shown by Welch here (it's proven in proof of Corollary 3.6)

After re-thinking my argument, I have realized where it fails. Namely, we cannot clock things like $\omega_1^2$, so we can't know if the simulation of machine with $0^\blacktriangledown$ has already made enough steps so that it will never halt.

Yes, standard Turing machine built in GoL will break at limit stages. Even if we somehow could make it work for more than $\omega$ steps, I suspect it's impossible to extend it past $\omega^2$ for this reason: if we could somehow simulate ITTM, we have to somehow know that it's limit stage. So there'd have to be some "check if it's limit" circuit, which would be active infinitely many times before $\omega^2$, which would probably break it.

Thank you a lot! Actually, I have found Welch's paper in which this result, along with lemma you used, is proved. Nevertheless, I appreciate your answer, as it's very nice and clear, and your argument is a lot easier to understand.