The result follows with $M=|a_n|$ actually as below:

Note that if $P$ has all the roots $w_j$ on the unit circle it follows that $w_j=1/\bar w_j$ so there is $c \ne 0$ st $$P(z)=cz^n\overline {P(1/\bar z)}=c\sum_{k=0}^n \bar a_kz^{n-k} $$ hence $a_k=c \bar a_{n-k}$ and in particular since $a_n \ne 0$ we have $a_0 \ne 0, |c|=1$ so $|a_0|=|a_n|$

But $a_0+a_nz^n=\frac{1}{n}\sum_{\omega^n=1}f(wz)$, so $|a_0+a_nz^n| \le \max_{|\zeta|=1}|P(\zeta)|$ for all $|z|=1$ hence choosing $|z|=1$ st $|a_0+a_nz^n|=|a_0|+|a_n|=2|a_n|$ the result follows

Note also that we only used that $P$ is self inversive ($w_j$ root iff $1/\bar w_j$ root and no zero roots) and actually only that $|a_0|=|a_n|$ which is implied by self-inversivity.

The result follows with $M=|a_n|$ actually as below:

Note that if $P$ has all the roots $w_j$ on the unit circle it follows that $w_j=1/\bar w_j$ so there is $c \ne 0$ st $$P(z)=cz^n\overline {P(1/\bar z)}=c\sum_{k=0}^n \bar a_kz^{n-k} $$ hence $a_k=c \bar a_{n-k}$ and in particular since $a_n \ne 0$ we have $a_0 \ne 0, |c|=1$ so $|a_0|=|a_n|$

But $a_0+a_nz^n=\frac{1}{n}\sum_{\omega^n=1}P(wz)$, so $|a_0+a_nz^n| \le \max_{|\zeta|=1}|P(\zeta)|$ for all $|z|=1$ hence choosing $|z|=1$ st $|a_0+a_nz^n|=|a_0|+|a_n|=2|a_n|$ the result follows

Note also that we only used that $P$ is self inversive ($w_j$ root iff $1/\bar w_j$ root and no zero roots) and actually only that $|a_0|=|a_n|$ which is implied by self-inversivity.