The answer is YES!

Let $\mathcal H_\alpha$ stand for the set of all sets hereditarily strictly subnumerous to ordinal $\alpha$.

Now for any set $x$, $\mathcal H^x_{min}$ is meant to be the minimal $\mathcal H_\alpha$ such that there exists an iterative power of it that is supernumerous to $x$. Formally:

Define: $\mathcal H^x_{min} = min \ \mathcal H_\alpha: \exists \beta [ x \rightarrowtail P_\beta(\mathcal H_\alpha)]$

Where $``\rightarrowtail"$ signify "*is injective to*".

$P_\beta$ is defined recursively as:

$P_\emptyset(x)=x \\ P_{\beta+1}(x)= P(P_\beta(x)) \\ P_\beta(x) = \bigcup (\{P_\alpha(x):\alpha < \beta\}) \text{ if } \beta \text{ is a limit ordinal} $

Now by $P^x_{min}(S) $ its meant the minimal iterative power of $S$ that is supernumerous to $x$. Formally:

Define: $P^x_{min} (S) = min \ P_\beta (S): x \rightarrowtail P_\beta(S) $

Now we come to define *cardinality* of a set $x$, denoted by $``|x|"$, as the set of all subsets of $P^x_{min} (\mathcal H^x_{min})$ , that are equinumerous to $x$. Formally:

Define: $|x|= \{y| \ y \sim x \land y \subseteq P^x_{min} (\mathcal H^x_{min}) \}$

Where $\sim$ signify "*is bijective to*"

This definition of cardinality can work under grounds weaker that those of Scott's cardinality? The latter demands the assumption that "*every set is equinumerous to some well founded set*", and under that assumption the cardinality [defined here] of any set would be exactly its Scott's cardinal. But this definition can work even when the above assumption fails, but it requires the statement: $$\forall x \exists \alpha \exists \beta : x \rightarrowtail P_\beta (\mathcal H_\alpha)$$ which doesn't imply the above assumption! (See this answer, and this).

Note: by *well founded set* its meant a set whose transitive closure is well founded with respect to $\in$.

beforeone knows the answer, then iflaterit turns out the asker gets the answer they should answer their own question. Asking and answering in one go is so uncommon, though, that this is the only example I can recall in a decade of using the site. $\endgroup$ – David Roberts yesterday