Why are these parabolic subgroups opposed? I am reading notes of Michel Brion on spherical varieties.
Consider a reductive group $G$, a Borel $B$ in $G$, a finite dimensional $G$-module $M$ and a closed orbit $Y$ of $G$ in $\mathbb{P}(M)$. The stabilisator of every point of $Y$ is a parabolic subgroup of $G$, so we can fix a $y \in Y$ such that $By$ is open in $Y$ (by Bruhat-decomposition). Also choose $m \in M$ such that $y=[m]$.
My Question is first of all: why is there an $\eta \in (M^*)^{(B)}$ such that $\eta(m)=1$. And why are the parabolic groups $G_{[\eta]}$ and $G_y$ opposed?
 A: Let's start with $x \in Y$ a $B$-fixed vector. Then it's a high weight line in $M$, so the linear span $N \leq M$ of $Y$ is an irreducible subrepresentation of $M$. 
Do you mind if I assume we're in characteristic $0$, as Brion does? Then we can split $M = N \oplus N'$, and $M^* = N^* \oplus N'^*$.` 
(Which is to say, you can reduce to the case $M$ irreducible, if that helps understand the situation.) So it's sufficient (and in the general case, necessary!) to look for your $\eta$ inside $N^*$.
Your first question is then why the $\eta \in (N^*)^B$, unique up to scale, has $\eta(m) \neq 0$. If it were zero,
$$ 0 = \eta(m) = (b\cdot \eta)(m) = \eta(b^{-1}\cdot m) \qquad \forall b\in B $$
from which we learn that $\eta$ annihilates a spanning set of $N$, so all of $N$. But $\eta \in N^*$, ` so it's zero, contradiction.
I'm not used to describing parabolic subgroups as "opposed" but I assume it means that $G_{[\eta]} G_y$ is dense in $G$. By definition of $\eta$, $G_{[\eta]} \geq B$, so it's enough to show $B G_y$ is dense in $G$, or that $B G_y / G_y$ is dense in $G/G_y$. This was exactly the condition used to pick $y$, that $B y$ was dense in $Y \cong G/G_y$.
A: Maybe it's helpful to add some comments to what Allen has already said, since the question is formulated loosely and can be looked at from different viewpoints.
1) It would help to point out explicitly which lecture notes by Brion you are looking at, since he has written and lectured a lot on spherical varieties.  In any case,  his standard framework involves a connected semisimple (or more generally, reductive) algebraic group over an algebraically closed field of characteristic 0 which is sometimes taken to be $\mathbb{C}$.   These assumptions are usually necessary for his use of classical finite dimensional representation theory, in which modules are completely reducible and the (absolutely) irreducible ones are parametrized by their dominant integral highest weights $\lambda$ relative to a fixed Borel subgroup $B$ and maximal torus $T$.    
2) As Allen observes, your module $M$ might as well be assumed irreducible, so it is generated by a highest weight vector.   The line through this highest weight vector has some standard parabolic subgroup $P \supset B$ as its stabilizer, so after passing to projective space you get a fixed point. 
3) How does the opposite parabolic subgroup $P^-$ come into the picture?   (This notion of opposite is standard in the literature and refers to a parabolic subgroup of $G$ whose intersection with $P$ is a Levi subgroup of each; the respective unipotent radicals then involve root groups for the "leftover" positive roots and their negatives respectively.)    The dual space $M^*$ also has the structure of an irreducible representation, but with highest weight $-w_\circ \lambda$ where $w_\circ$ is the longest element of the Weyl group.  Here the lowest weight space of weight $-\lambda$ has stabilizer $P^-$.  And a lowest weight vector in the dual space can be chosen to be a linear functional with value 1 at the given highest weight vector of $M$.  This should be basically what you are looking for, in the projectized version used by Brion.
The formalism here is elementary but complicated.   It probably goes back to Chevalley's standard construction of irreducible modules inside the algebra of regular functions on $G$.   (There are various textbook accounts, including section 31 of my book.  See especially the paragraph at the top of page 194.) 
