Expanding what I wrote in a comment.
First, cardinality is hard to define in constructive math. We certainly get an equivalence relation (ignoring the issue of exactly what this is an equivalence relation on) by considering the notion “$X$ and $Y$ are in bijection”, and is tempting to write this as $|X|=|Y|$. Now classically we can define a partial order relation $|X|\leq|Y|$ by “$X$ injects into $Y$” (which, in ZFC, turns out to even be a total order); but the fact that this is an order, or more precisely, that this is antisymmetric (if $X$ injects into $Y$ and $Y$ injects into $X$, then there is a bijection between $X$ and $Y$) is the Cantor-Bernstein theorem, which is not valid constructively. Instead, “$X$ injects into $Y$” just gives us a preorder relation, and now we have two different equivalence relations which might define a form of cardinality: “$X$ and $Y$ are in bijection” and the coarser “$X$ injects into $Y$ and $Y$ injects into $X$” equivalence relation obtained from this preorder. I don't know of much study of these relations, but they are probably not very well behaved: for example, with respect to surjections, already in classical ZF, a set might surject onto a set of strictly larger cardinality; maybe there is a notion to be made of cardinality from surjections, but already in ZF this does not behave very well, so I'm pessimistic.
Nevertheless, some notions are fairly standard (although the names below aren't universal):
A set is said to be finite when it is in bijection with $\{1,\ldots,n\}$ for a certain natural number $n$ (possibly $n=0$ with the convention that $\{1,\ldots,n\} = \varnothing$ when $n=0$; replace $\{1,\ldots,n\}$ by $\{0,\ldots,n-1\}$ if you prefer numbering things from $0$ so that it works better; anyway, this isn't relevant here because, constructively, natural numbers are discrete: if $i,j\in\mathbb{N}$ then $i<j$ or $i=j$ or $i>j$ holds). We can then call $n$ (which is well-defined) the cardinal of the set, and this notion behaves well, but, of course, it is defined only for some very specific sets.
A set is said to be finitely enumerated (or “Kuratowski-finite”, and some people just call this “finite”) when it is surjected by $\{1,\ldots,n\}$ for some natural number $n$. Note that you might be tempted to ask about the smallest $n$ such that $\{1,\ldots,n\}$ surjects the set, but this need not exist (even if an $n$ does exist, i.e., the set is finitely enumerated), so no notion of cardinality from this.
A set is said to be subfinite when it injects into $\{1,\ldots,n\}$ for some natural number $n$, i.e., the set is in bijection with a subset of some $\{1,\ldots,n\}$ (equivalently, a bounded subset of $\mathbb{N}$), i.e., it is a subset of a finite set. Again, you might be tempted to ask about the smallest $n$ such that this holds, but it need not exist (even if the set is subfinite).
A set is said to be subfinitely enumerated when it is surjected by a subset of $\{1,\ldots,n\}$ for some natural number $n$. It is also true that this is the same as being a subset of a finitely enumerated set, although this is not completely obvious.
The first two notions are clearly more important than the other two, but all four arise. Also, “finite” is equivalent to “finitely enumerated and discrete” and to “finitely enumerated and subfinite”. (A set $X$ is “discrete” when $(x=y)\lor\neg(x=y)$ for all $x,y\in X$.)
We can also make the same definitions replacing $\{1,\ldots,n\}$ by $\mathbb{N}$. Terminology is… all over the place. I suppose it's more reasonable to call them something like “exactly countable”, “countably enumerated”, “subcountable” and “subcountably enumerated”, but here maybe you also want to add notions which replace $\mathbb{N}$ by a decidable subset of $\mathbb{N}$ (a subset $S\subseteq X$ is said to be “decidable” when $(x\in S)\lor\neg(x\in S)$ for all $x\in X$).
Lost? That's pretty much the reason people don't look at cardinality in constructive math: it quickly descends into a myriad of hard-to-relate notions.
Nevertheless, what can be said about $\Omega := \mathcal{P}(1)$?
Well, I should first point out that I'm implicitly working in something like IZF or the internal logic of a topos. In stronger (“even more constructive”, as one might call them, or perhaps more accurately “more predicative”) theories like CZF, $\Omega$ might… not even exist. And this is arguably because it might be too large to be a set. But let's stick to a framework in which we're allowed to take powersets.
Here's one thing:
Proposition: If $\Omega$ is discrete, then Excluded Middle holds. In particular, if $\Omega$ is finite or even subfinite (“subfinite” implies “discrete”), then Excluded Middle holds, and $\Omega$ is, in fact, finite with cardinality $2$.
Proof. That a set $X$ is “discrete” means that $(x=y)\lor\neg(x=y)$ for all $x,y\in X$. In the case of $\Omega$, equality is equivalence, so we are asserting $(p\Leftrightarrow q)\lor\neg(p\Leftrightarrow q)$ for all $p,q\in\Omega$. Applying this to $q=\top$ and using the fact that $(p\Leftrightarrow\top)$ is just $p$, we get $p\lor\neg p$ for all $p\in\Omega$. This is LEM. And of course, if LEM holds, then we are in classical logic and it is trivial that $\Omega$ has exactly two elements. ∎
Along similar lines:
Proposition: (A) If $\{1,\ldots,n\}$ injects into $\Omega$ then $n\leq 2$. ❧ (B) If $\Omega$ surjects onto $\{1,\ldots,n\}$ then $n\leq 2$.
Proof. For (A), it is enough to show that $\{1,2,3\}$ does not inject into $\Omega$ (because $\neg(n\geq 3)$ implies $n\leq 2$ for natural numbers). If it does, then there are $p_1,p_2,p_3\in\Omega$ such that $\neg(p_1\Leftrightarrow p_2)$ and $\neg(p_2\Leftrightarrow p_3)$ and $\neg(p_3\Leftrightarrow p_1)$. But similarly, for (B), it is enough to show that $\Omega$ does not surject onto $\{1,2,3\}$ (because if it surjects onto $\{1,\ldots,n\}$ for $n\geq 3$ then it certainly surjects onto $\{1,2,3\}$), and this again gives us $p_1,p_2,p_3$ with the same conditions.
So we just have to show a contradiction from the existence of $p_1,p_2,p_3\in\Omega$ such that $\neg(p_1\Leftrightarrow p_2)$ and $\neg(p_2\Leftrightarrow p_3)$ and $\neg(p_3\Leftrightarrow p_1)$.
Now note that any of the eight “classical” combinations $p_1\land p_2\land p_3$, $p_1\land p_2\land\neg p_3$ through $\neg p_1\land\neg p_2\land\neg p_3$ trivially gives a contradiction. So the negation of the disjunction $(p_1\land p_2\land p_3) \lor \cdots \lor (\neg p_1\land\neg p_2\land\neg p_3)$ holds. But by distributivity, this eightfold disjunction is just $(p_1\lor\neg p_1) \land (p_2\lor\neg p_2) \land (p_3\lor\neg p_3)$. So the disjunction of that holds. But we have $\neg\neg(p_1\lor\neg p_1)$ and similarly or the other two. So $\neg\neg((p_1\lor\neg p_1) \land (p_2\lor\neg p_2) \land (p_3\lor\neg p_3))$ (using the standard fact that $\neg\neg$ distributes over conjunctions). But we just proved $\neg((p_1\lor\neg p_1) \land (p_2\lor\neg p_2) \land (p_3\lor\neg p_3))$. So we have a contradiction. ∎
This shows that in a certain sense, $\Omega$ can't be too large. Note that $\{1,2\}$ (OK, I really should have started numbering at $0$ and call this $\{0,1\}$) does inject into $\Omega$ as $\{\bot,\top\}$, so “the largest $n$ such that $\{1,\ldots,n\}$ injects into $\Omega$” is well-defined and has the value $2$ exactly.
I might as well mention:
Proposition: If $\Omega$ surjects onto $\{0,1\}$ (equivalently, $\{1,2\}$) then the Weak Law of Excluded Middle (that is $\neg p\lor\neg\neg p$ for all $p\in\Omega$ holds), and conversely.
Proof. We assume we have a function $\chi\colon\Omega\to\{0,1\}$ surjective. Since we can replace $\chi$ by $1-\chi$, we might as well assume $\chi(\top)=1$. By surjectivity, there is $p\in\Omega$ such that $\chi(p)=0$. Now $p=\top$ is an obvious contradiction, so $\neg(p=\top)$; but $p=\top$ means the same as $p$, so $\neg p$; so $p=\bot$; so now we have $\chi(\bot)=0$.
If $p\in\Omega$ is such that $\chi(p)=0$ then $p=\top$ is a contradiction, so $\neg p$ (same reasoning as above). Similarly, if $\chi(p)=1$ then $p=\bot$ is a contradiction, so $\neg\neg p$. But since we must have $\chi(p)=0$ or $\chi(p)=1$, we have $\neg p$ or $\neg\neg p$, which is WLEM.
As for the converse: if $\neg p\lor\neg\neg p$ holds for all $p\in\Omega$, we can define $\chi(p)=0$ when $\neg p$ and $\chi(p)=1$ when $\neg\neg p$, and WLEM ensures that this $\chi$ is defined on all of $\Omega$, and it is surjective since $\chi(\bot)=0$ and $\chi(\top)=1$. ∎
All these things tend to suggest that $\Omega$ can't be “large” in the sense of containing a large finite set (let alone $\mathbb{N}$) or surjecting onto one.
Now on the other hand, $\Omega$ can be “large” in the following sense:
(Meta-)Proposition: It is consistent that no subset of $\mathbb{N}$ surjects onto $\Omega$.
Proof. Working externally in a classical universe, consider the topos of sheaves over $\mathbb{R}$. Since an injection of sheaves gives an injection of stalks at a point and a sujection of sheaves gives a surjection of stalks at a point, any sheaf which is internally surjected by a subset of the natural numbers object (viꝫ. the constant sheaf with value $\mathbb{N}$) has countable stalk at any point. But $\Omega$ is the sheaf of open sets of $\mathbb{R}$, and it is easy to see that there are more than countably many germs of open sets at a point. ∎
(Technical note: this is a topos-based argument; if you want a consistency result with IZF, construct the cumulative hierarchy within the topos in question.)
TL;DR: There is no satisfactory notion of cardinality; nevertheless, $\Omega$ can't be large in the sense that it can't contain or surject onto $\{1,\ldots,n\}$ unless $n\leq 2$; however, you might not be able to surject a subset of $\mathbb{N}$ onto it, meaning it might fail to be “countable” even in the weakest possible sense of the word (“subcountably enumerated” or whatever you might want to call it).
