Let angles between rays is equal $\frac{2\pi}{3}$ Let $PP_1=x,PP_2=y,PP_3=z$ We will show that $x+y+z \leq 3r$ For triangle $P_1P_2P_3$ we have that its circumradius is $r$ and its sides are $\sqrt{x^2+xy+y^2},\sqrt{y^2+yz+z^2},\sqrt{z^2+zx+x^2}$ and its area is $ \frac{\sqrt{3}}{4}(xy+yz+zx)$ We have $\frac{\sqrt{x^2+xy+y^2}\sqrt{y^2+yz+z^2}\sqrt{z^2+zx+x^2}}{\sqrt{3}(xy+yz+zx)}=r$ So enough to prove that $$\frac{ \sqrt{3}\sqrt{x^2+xy+y^2}\sqrt{y^2+yz+z^2}\sqrt{z^2+zx+x^2}}{xy+yz+zx} \geq x+y+z$$ $x^2+xy+y^2 \geq \frac{3}{4}(x+y)^2$ $\sqrt{3}\sqrt{x^2+xy+y^2}\sqrt{y^2+yz+z^2}\sqrt{z^2+zx+x^2} \geq \frac{9}{8}(x+y)(x+z)(y+z)=\frac{9}{8}((x+y+z)(xy+yz+zx)-xyz)=(x+y+z)(xy+yz+zx) +\frac{(x+y+z)(xy+yz+zx)-9xyz}{8} \geq (x+y+z)(xy+yz+zx) $ And so for such choice of rays we will have \[|PP_1|+|PP_2|+|PP_3|\leq 3r.\]

Actually just Ragvaldo's initial idea is enough. Let angles between rays be $\frac{2\pi}{3}$. Then point P is the Fermat point of triangle $P_1P_2P_3$ and $r$ is the circumradius of the same triangle. It is well known that Fermat point is the point for which the sum of lengths from the vertices of the triangle is minimized so it is no larger than the sum of the distances from the vertices of the circumcenter (which is equal to $3r$). Thus \[|PP_1|+|PP_2|+|PP_3|\leq 3r.\]