In Chapter 15 at the bottom of page 228 of $\textit{Set Theory}$ by Jech, he writes that if $\kappa$ is a cardinal in $V$ and if $\kappa$ has no new bounded subsets in $V[G]$, then $\kappa$ remains a cardinal in $V[G]$. I do not see why this holds.
In Jech's particular example, this property follows from the fact that $P$ is $<\kappa$-closed. I can see that if $P$ is $<\kappa$-closed then all cardinals less than or equal to $\kappa$ is preserve. However I am using $\kappa$-closed rather than just that no new bounded subsets are adding: if $\lambda \leq \kappa$ and $V \models \lambda$ is a cardinal. If $V[G] \models \lambda$ is not a cardinal, then there exists a $f : \alpha \rightarrow \lambda$ where $\alpha < \lambda$ and $f$ is a bijection. By $<\kappa$-closed, $f \in V$. By absoluteness, $f$ is a bijection betweeen $\lambda$ and $\alpha < \lambda$ in $V$. This contradicts $\lambda$ being a cardinal in $V$.
In general, if no bounded subsets of a cardinal in $V$ is added in $V[G]$, how does one see that $\kappa$ remains a cardinal in $V[G]$?
Suppose that $\kappa$ is no longer a cardinal, then there is some cardinal $\alpha<\kappa$ and $f\colon\alpha\to\kappa$ which is a bijection. In particular, this means that there is a subset of $\alpha\times\alpha$ which encodes a well-ordering of order type $\kappa$.
Since we can canonically identify $\alpha\times\alpha$ with $\alpha$, it follows that there is a new subset of $\alpha$ which encodes a well-ordering of order type $\kappa$. This subset is also a subset of $\kappa$, but it is a bounded subset of $\kappa$, since $\alpha$ is its bound.
So if no bounded subsets were added to $\kappa$, then not only that $\kappa$ is still a cardinal, the cardinal structure (cofinalities and whatnot) up to $\kappa$ is preserved (by the same argument).