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 phenomonon 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.
First, let me explicitly assume $R$ is a noetherian local domain with a principal maximal ideal $\mathfrak{m}$.
Proposition. The Krull dimension of $R$ is at most $1$.
Proof. Let $\mathfrak{m} = (t)$, and let $\mathfrak{p}$ be prime. We know $\mathfrak{p} \subseteq \mathfrak{m}$, so it is enough to show that either $\mathfrak{p} = \mathfrak{m}$ or $\mathfrak{p} = (0)$. Suppose $\mathfrak{p} \ne \mathfrak{m}$: then $t \notin \mathfrak{p}$. Let $a_0 \in \mathfrak{p}$. For each $a_n$, because $\mathfrak{p}$ is prime, there exists $a_{n+1}$ in $\mathfrak{p}$ such that $a_n = a_{n+1} t$. By the axiom of dependent choice, this yields an infinite ascending sequence of principal ideals:
$$(a_0) \subseteq (a_1) \subseteq (a_2) \subseteq \cdots$$
Since $R$ is noetherian, for $n \gg 0$, we must have $(a_n) = (a_{n+1}) = (a_{n+2}) = \cdots$. Suppose, for a contradiction, that $a_0 \ne 0$. Then, $a_n \ne 0$ and $a_{n+1} \ne 0$, and there is $u \ne 0$ such that $a_{n+1} = a_n u$. But then $a_n = a_{n+1} t = a_n u t$, so cancelling $a_n$ (which we can do because $R$ is an integral domain), we get $1 = u t$, i.e. $t$ is a unit. But then $\mathfrak{m} = R$ – a contradiction. So $a_n = 0$. $\qquad \blacksquare$
Here's an elementary proof which shows why we can reduce to the case where $R$ is an integral domain.
Proposition. Any non-trivial ring $A$ has a minimal prime.
Proof. By Krull's theorem, $A$ has a maximal ideal, which is prime. Let $\Sigma$ be the set of all prime ideals of $A$, partially ordered by inclusion. The intersection of a decereasing chain of prime ideals is a prime ideal, so by Zorn's lemma, $\Sigma$ has a minimal element. $\qquad \blacksquare$
Thus, we can always assume that a maximal chain of prime ideals starts at a minimal prime and ends at a maximal ideal. But if $R$ is a noetherian local ring with principal maximal ideal $\mathfrak{m}$ and $\mathfrak{p}$ is a minimal prime of $R$, then $R / \mathfrak{p}$ is a noetherian local domain with a principal maximal ideal $\mathfrak{m}$, and $\dim R = \sup_\mathfrak{p} \dim R / \mathfrak{p}$, as $\mathfrak{p}$ varies over the minimal primes.
Update. Georges Elencwajg pointed out in a comment that the first proof actually works without the assumption that $R$ is a domain, because $(1 - u t)$ is always invertible.
Best Answer
I assume you are talking about the completion of a local Noetherian ring $A$ with respect to the topology induced by its maximal ideal $m$. Then $\hat{A}$ is again Noetherian local ring with maximal ideal $m \hat{A}$. Reference: Matsumura's Commutative Ring Theory p. 63.