Show that given a point $P$ and a hyperplane $H \subseteq \mathbb{P}^n$ such that $P \in H$, there is a linear transformation $T$ such that $T(P)=(0:\cdots:0:1)$ and $H$ is given by the equation $X_0=0$.
A hyperplane in $\mathbb{P}^n$ has equation $a_0X_0+\cdots+a_nX_n=0$.
Let $P=(b_0:\cdots:b_n)$.
Since $P \in H$, $a_0b_0+\cdots+a_nb_n=0$.
How can I proceed? Some help would be appreciated. Thanks.
Go up one dimension and into affine space. There $P$ corresponds to the line $L = (tb_0, tb_1,\ldots,tb_n)$ and $H$ still has the same equation.
Now apply a linear transformation to $\Bbb A^{n+1}$ that moves $H$ to be orthogonal to $(1, 0,0,\ldots,0)$ (remember that $H$ originally is orthogonal to $(a_0, \ldots,a_n)$) (this might move $L$ to someplace else). Then apply another linear transformation that fixes $(1, 0,\ldots,0)$, but rotates the (new) $L$ to the line $(0,0,\ldots,0,s)$. Lastly, project back down to $\Bbb P^n$.
I don't know that this gives you the neatest description of the transformation, but it is, at least in my opinion, the most straight-forward way to prove that it is possible to do it.