Suppose I want to prove that $P_2$ is isomorphic to $R^3$. Why is it not sufficient to prove dim($P_2$)=dim($R^3$) and apply Thm part (b)? Why do I have to show that the ker($T$)=$0$?
There is a difference between "two spaces are isomorphic" and "the linear transformation $T$ is an isomorphism between the two spaces".
If have two (finite dimensional) vector spaces $V$ and $W$, and you want to check that two are isomorphic, it's enough to compare dimensions. If you also have a linear transformation $T:V\to W$ ($V, W$ still finite dimensional), and you want to check that $T$ is an isomorphism, then you have to check that $V$ and $W$ are isomorphic (i.e. compare dimensions) and also that $T$ is injective (i.e. check the kernel).
The theorem you provided is all about the latter case, while it seems that your actual problem (comparing $P_2$ to $\Bbb R^2$) is about the former.