 Solution

The generic extension is $M[G]=\{{\tau }^G:\tau \in M\text{ is a forcing name}\}$. It is transitive: if $x\in {\tau }^G$, an active pair in $\tau$ supplies a subname $\sigma \in M$ with $x={\sigma }^G$. Ground sets belong to it by their canonical names. Without assuming a weakest condition, use $\check{a}=\{(\check{b},p):b\in a,p\in P\}$; nonemptiness of $G$ gives ${\check{a}}^G=a$.

Induction on forcing name rank gives $\mathrm{rank}({\tau }^G)\le \mathrm{nrk}(\tau )$. The right side is an ordinal of $M$. If $\alpha \in M[G]$ is an ordinal, its rank equals $\alpha$, so it is at most a ground-model ordinal. Transitivity of $M$ then implies $\alpha \in M$. Conversely every ground ordinal remains the same ordinal in the transitive extension, since membership is unchanged. HenceThis proves forcing preserves ordinals directly from ranks, without assuming cardinal preservation.