A set equinumerous with some ordinal has a least such ordinal, that ordinal is a cardinal, and equinumerous sets get the same one; no choice principle is used

If $<$ well-orders $X$ then $(X, <)$ has an order type $\alpha$ and the collapsing map is an order isomorphism, hence a bijection $X \to \alpha$, so $X \approx \alpha$.

Conversely, if $f : X \to \alpha$ is a bijection then $x <_X y :\iff f(x) \in f(y)$ is a well-order of $X$, since $f$ transports irreflexivity, transitivity, trichotomy and the least-element property of $\in$ on $\alpha$ back to $X$; this proves claim (a).

Assume now $X \approx \alpha$ for an ordinal $\alpha$, and put $C = \{\xi \in \alpha^{+} : \xi \approx X\}$, a set by Separation inside the ordinal $\alpha^{+}$, all of whose elements are ordinals, and nonempty because $\alpha \in \alpha^{+}$ and $\alpha \approx X$.

By [L4] the set $C$ has an $\in$-least element $\kappa$, and $\kappa \approx X$.

$\kappa$ is least among all ordinals equinumerous with $X$: given $\beta \approx X$, trichotomy gives $\beta \in \alpha^{+}$, in which case $\beta \in C$ and $\kappa \subseteq \beta$ by minimality; or else $\alpha^{+} \subseteq \beta$, in which case $\alpha \in \alpha^{+} \subseteq \beta$ gives $\alpha \subseteq \beta$, while $\alpha \in C$ gives $\kappa \subseteq \alpha$, so again $\kappa \subseteq \beta$. This proves claim (b), with $\lvert X \rvert := \kappa$.

Claim (c): if $\gamma \in \kappa$ had $\gamma \approx \kappa$ then $\gamma \approx X$ by [L5], so $\kappa \subseteq \gamma$ by step 3.1, whence $\gamma \in \kappa \subseteq \gamma$ and $\gamma \in \gamma$, which [L3] forbids; so $\kappa$ is a cardinal.

Claim (d): if $X \approx Y$ then an ordinal is equinumerous with $X$ exactly when it is equinumerous with $Y$, by symmetry and transitivity of $\approx$, so the two least such ordinals coincide; and $Y \approx \kappa$ makes $Y$ well-orderable by step 1.2.

Claim (e): $\alpha \approx \alpha$, so $\lvert \alpha \rvert \subseteq \alpha$ by step 3.1; if $\alpha$ is a cardinal then no $\xi \in \alpha$ is equinumerous with $\alpha$, so the least ordinal equinumerous with $\alpha$ is $\alpha$ itself; and conversely $\lvert \alpha \rvert = \alpha$ makes $\alpha$ a cardinal by step 4.1.

Assuming the Axiom of Choice, every set $X$ carries a well-order by [L7], hence $X \approx \lvert X \rvert$ by step 1.1 and $\lvert X \rvert$ is defined for every set; and by step 3.1 it is the least ordinal equinumerous with $X$, which is what [L6] calls the cardinality of $X$.