You want to find a projection $P$ (that is, a linear transformation $P$ such that $P^2=P$) which is not zero, and such that $AP=0$. But you need to assume that there is a $\mathbf{v}\neq \mathbf{0}$ such that $A\mathbf{v}=\mathbf{0}$. Otherwise, $A$ is invertible, and then $AP=0$ implies $P=0$.
Correction. In context: we have a group $G$, and two inequivalent irreducible representations $D_1$ and $D_2$ of $G$ such that $D_1(g)A = AD_2(g)$ for all $g\in G$. The desired conclusion is to show that $A=0$.
"The subspace that annihilates $A$ on the right" is the kernel of $A$: the collection of all vectors $\mathbf{v}$ such that $A\mathbf{v}=\mathbf{0}$. You want the projection onto the kernel of $A$; there is an implicit assumption that $\mathbf{v}\neq\mathbf{0}$, as the next paragraph argues that if $A$ does not have any vector that annihilates it on either side, then $A$ must be invertible.