Let us consider a model of ZFC $M$, a forcing $\mathbb{P}\in M$. If $G$ is a $\mathbb{P}$-generic filter over $M$ we can construct the generic extension $M[G]$ of $M$. Given any cardinal number $\kappa$ in $M[G]$ it is clear $\kappa$ is a cardinal in $M$ since $M$ cannot since more bijections than $M[G]$.
My question is, suppose for example $\omega_1^M$ is a cardinal number in $M[G]$, then $\omega_1^M=\omega_1^{M[G]}$? And in general, if $\kappa^M$ is a cardinal in $M[G]$ then $\kappa^M=\kappa^{M[G]}$?
Thanks in advance.
You can easily check that $\omega_1^M\le \omega_1^{M[G]}$ holds as an ordinal. Moreover, since $\omega_1^M$ is cardinal in $M[G]$, $\omega_1^M$ must be an uncountable ordinal. Thus $M[G]\models \omega_1^{M[G]}\le \omega_1^M$ so they are equal. (It follows from the fact that if $\alpha$ is uncountable ordinal then $\alpha\ge\omega_1$. Take $\alpha=\omega_1^M$ and relativize the formula to $M[G]$.)
I don't know the meaning of $\kappa^M$, because $\kappa^M$ is just $\kappa$. However if you have intended such as $\omega_2^M$, it does not hold in general. For example, you can collapse $\omega_1^M$ to a countable ordinal and preserve cardinals above $\omega_1^M$. In that case $\omega_2^M = \omega_1^{M[G]}$ so $\omega_2^M$ is cardinal in $M[G]$ but $\omega_2^M\neq \omega_2^{M[G]}$.