@TomLeinster I think what I wrote is ok. The functor sends a monad to the corresponding free functor into the category of algebras which is a left adjoint under $\mathcal{C}$. Perhaps the confusion stems from the fact that my adjunction is covariant whereas in the result you cite it is contravariant.

I might have misinterpreted but is there a claim here that every $p$-complete space is the limit of some $p$-profinite space? I thought this might fail for some wild enough non-nilpotent spaces, I didn't know mandell theorem was that general...

Also in what sense does computing $GX$ for finite $X$ solves the word problem for groups? Presumably you could calculate $GX$ but still find it very hard to compare to $GY$ for any other $Y$. Similarly to how difficult it is to compare arbitrary Kan complexes to each other. Most likely you meant something more subtle which I missed.

It could still be possible that such a fibrant replacement functor exists for the (co-)cartesian model structure on marked simplicial set right? That is, one that preserves finite limits and fibrations.

Yeah, me too. That's why I asked.

Do you happen to know what's the analogous statement for $\infty$-categories?

For instance when $p>2$ and $n$ is odd and $1<k<n$ then $K(\mathbb{Z}/p,n-k)$ is kind of a cartier dual of $K(\mathbb{Z}/p,k)$ in the $K(n)$-local category (at least at the level of dieudonne modules it becomes actually true - to make this work spectrally you need to set up the correct framework which I don't remember how to do currently). In particular they will not be self (cartier) dual.

I feel like many times the interesting notion of duality for hopf algebras is different than the symmetric monoidal one. I'm thinking of cartier duality and maybe even its chromatic variant for p-finite eilenberg maclane spaces in the $K(n)$-local category.

Isn't the $L_{K(n)} K(\mathbb{Z}/p,n)$ being a hopf algebra kind of a red heering here? In any $\infty$-semiadditive symmetric monoidal category all $\pi$-finite spaces will be self-dual.