 Solution

The needed general-position theorem is sharp triple transitivity of the projective linear group on the projective line: for any two ordered triples of distinct projective points, exactly one projective transformation takes the first triple to the second.

In particular, let $T_P$ take $(P_0,P_1,P_2)$ to $(\infty ,0,1)$. In these coordinates,so cross-ratio invariance gives $T_P(P_3)=\lambda$, where $\lambda =(P_0{P}_1:P_2{P}_3)$. Distinctness implies $\lambda \ne 0,1$ and $\lambda$ is finite. The entire ordered quadruple is thus normalized toFor a second quadruple $(Q_0,Q_1,Q_2,Q_3)$ with the same cross-ratio, its normalizing map $T_Q$ yields the same quadruple. ConsequentlyThis proves the equivalence. Uniqueness on the first three points also shows that the projectivity carrying one ordered quadruple to the other is unique. The cross-ratio normalization of four projective points therefore gives a complete invariant for this ordered classification.