Since $ P_0 = Q_0 $ , P and Q have the same roots, and so much be multiples of each other. So let $ P(z) = k Q(z) $. $ P_1 = Q_1 $, so there exists a complex number $ a $ such that $ P(a) = Q(a) = 1 $. But $ P(a) = k Q(a) $, thus $ k=1 $, and $ P(z) = Q(z) $.

Gib Z wrote: Since $ P_0 = Q_0 $ , P and Q have the same roots, and so much be multiples of each other. .... Why ? $P(z)=z(z-1)^2$ and $Q(z)=z^2(z-1)$ are such that $P_0=Q_0$ and are not multiples of each other ...

Gib Z wrote: Since $ P_0 = Q_0 $ , P and Q have the same roots, and so much be multiples of each other. So let $ P(z) = k Q(z) $. $ P_1 = Q_1 $, so there exists a complex number $ a $ such that $ P(a) = Q(a) = 1 $. But $ P(a) = k Q(a) $, thus $ k=1 $, and $ P(z) = Q(z) $. The difficult part of the problem is that they may have different multiplicities for each root and simply you have ignored that. The condition $P_1=Q_1$ was possibly given to tackle this issue.

Let $k=|P_0|=|Q_0|$ and $l=|P_1|=|Q_1|$. $P(z)-Q(z)=(Q(z)(P(z)-1))-(P(z)(Q(z)-1))$, so $deg(P(z)-Q(z)) \geq k+l$. By https://artofproblemsolving.com/community/c7h58520p357656 it follows that $k+l \geq n+1$, which is a contradiction since $deg(P(z)-Q(z)) \leq n$, where $n=max\{deg P, deg Q\}$.