= Solution
The needed general-position theorem is \b[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,
$$
(\infty\,0:1\,z)=z,
$$
so <cross-ratio> invariance gives $T_P(P_3)=\lambda$, where $\lambda=(P_0P_1:P_2P_3)$. Distinctness implies $\lambda\ne0,1$ and $\lambda$ is finite. The entire ordered quadruple is thus normalized to
$$
\boxed{(\infty,0,1,\lambda).}
$$
For 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. Consequently
$$
\boxed{T_Q^{-1}\circ T_P(P_i)=Q_i\quad(0\leq i\leq3).}
$$
This 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.
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