The linked MO question shows that the answer is yes when $\mathfrak p$ is maximal, the key point in that case being that the completion of $A$ at a maximal ideal is automatically local.
If $\mathfrak p$ is not maximal, then there will be a natural map
$\hat{A}_{\hat{\mathfrak p}} \to \widehat{A_{\mathfrak p}}$.
Indeed, there is a map $A \to A_{\mathfrak p}$, which induces a map of
$\mathfrak p$-adic completions
$\hat{A} \to \widehat{A_{\mathfrak p}},$
which in turn induces a map
$\hat{A}_{\hat{\mathfrak p}} \to \widehat{A_{\mathfrak p}}.$
But this map won't be an isomorphism if $\mathfrak p$ is not maximal
(at least if $A$ is Noetherian, so that the completions are reasonably
behaved).
To see why, consider as an example the case
when $A = \mathbb Z_p[x]$, and $\mathfrak p = (x)$.
Then $A_{\mathfrak p} = \mathbb Q_p[x]_{(x)},$ and so $\widehat{A_{\mathfrak p}}
= \mathbb Q_p[[x]].$
On the other hand, $\hat{A}_{\hat{\mathfrak p}} = \mathbb Z_p[[x]]_{(x)}$, which is a proper subring of $\mathbb Q_p[[x]]$. (E.g. an element of
$\mathbb Z_p[[x]]_{(x)}$ defines a meromorphic function on the $p$-adic disk
$1 > |x|$ with only finitely many zeroes and finitely many poles, none of
the latter being at $x = 0$; in particular, it has a non-trivial radius
of convergence around $0$. On the other hand, a typical function in $\mathbb Q_p[[x]]$
does not have any non-trivial radius of convergence around $0$.)
This illustrates the general phenomenon that when $\mathfrak p$ is not maximal,
so that $\hat{A}_{\hat{\mathfrak p}}$ is a genuinely non-trivial localization,
only certain $\mathfrak p$-adic limits exist in the localization
$\hat{A}_{\hat{\mathfrak p}}$, while by construction
$\widehat{A_{\mathfrak p}}$ is $\mathfrak p$-adically complete.
