Suppose we can map the circle to a polygon. We assume wlog $P$ maps $e^{it},0<t<\theta<\frac{\pi}2$ to $z_0+\lambda(z_1-z_0)$. Replace $P$ with $\frac{P-z_0}{z_1-z_0}$, we see $P(e^{it})$ is real when $0<t<\theta$. We can write the imaginary part of $P(\cos(t)+i\sin(t))$ as $g(t)=\sum a_n \cos(nt)+\sum b_n\sin(nt)$, where $a_n,b_n$ are reals. Then $g(t)=0,0<t<\theta$. we can further write $g(t)=0$ as a polynomial equation of $r=\tan(\frac t2)$. Or equivalently $h(r)=0,0<r<\tan(\frac{\theta}2)$. Obviously $h(r)=0,\forall r$, which implies $P(e^{it})$ is real $-\frac{\pi}2<t<\frac{\pi}2$. In other words, $P$ will map the half circle to one side. Similarly, $P$ will map the other half circle to another side. Hence, there is no way to map the entire circle to a polygon.