 Solution

Let $C(x)$ be the formula expressing cardinal number status:Ordinalhood and the assertion that a given set is a bijection between fixed sets are each expressible by a bounded formula in set theory. If $M$ regarded $x$ as a non-cardinal ordinal, a witnessing smaller ordinal and bijection would remain present and valid in $N$. Ordinalhood itself is absolute between the transitive models. Thus $N\models C(x)$ implies $M\models C(x)$: this is downward absoluteness of cardinalhood.

For failure in the other direction, put $\beta =({\omega }_1{)}^M$. This ordinal belongs to $M$, and $M\models C(\beta )$ with $\beta >\omega$. Externally $\beta$ is countable, since $\beta \subseteq M$ and $M$ is countable. By the construction of $N$, it contains a countability witness for $\beta$, and hence a bijection $\omega \to \beta$. Therefore $N\models \neg C(\beta )$. This formula is downward absolute and fails to be upward absolute.