Here, $| A |=3,| B |=4$
Let $\angle E O P=\phi$, then
$B =R \cos \phi=4$ and $A =R \sin \phi=3$
$\therefore \tan \phi=\frac{3}{4}$
Now, $\cos \phi=\frac{4}{5}$
Projection of $B$ on $O P$ is $| B | \cos \phi$
i.e. $\quad 4 \cos \phi=4 \times \frac{4}{5}=\frac{16}{5}=3.2$