I had been thinking of Grass$(p,r)$ as the set of $r$ dim subspaces of $k^p$ where $k$ is whatever ground field we take in the context. We then topologize this set.
Now in one of his papers, Nitsure starts by saying : Let $G = G(p,r)$ be the Grassmannian variety of $r$-dim quotients of $k^p$
He also goes on to say that we can think of the same as $SL(p)/P$ where $P$ is a parabolic subgroup.
Why does he call them quotients whereas in differential geometry we think of the elements as subspaces?
How do we get the second formulation and what is $P$?
This becomes becomes more evident, when you want to work with the projective space as a scheme instead of a manifold or abstract variety. Let's look at the easiest case, the projective line: When $k$ is a field, there is no difference between quotients and subspaces, since everything is split. When we want to look at $\mathbb{P}^1_k(R)$ for an $R$-algebra $k$ (really you can do this for any $R$-algebra), you might be tempted to define it via subspaces, but this turns out to be insufficient, whereas quotients yield a more easy description. I recommend you look at this answer. As a caveat it should be stressed, that also you have to switch from free modules to projective modules.
$P$ in this case is the $(r,p-r)$ parabolic. There are matrixes where the lower left $(p-r)\times r$ block is just $0$. The map goes as follows: Let's define $x\in G(p,r)$ to be the standard flag, i.e. $x=<e_1,...,e_r>$ with $e_i$ the standard basis. Given an element $g\in SL(p)$, we set $g.x:=<g(e_1),...,g(e_r)>$. You should verify that this defines a well-defined action. So you get a map $$ SL(p)\to Gr(r,p)$$ sending $g$ to $g.x$. This is surjective, because the action is transitive. Finally you verify that $Stab_{SL(p)}(x)=P$, so you get exactly that $$ SL(p)/P\to Gr(r,p)$$ is an isomorphism. This isomorphism in fact exists in greater generality for any parabolic subgroups (don't worry if you don't know what that means). You can consult these notes for more details.