= Solution
{c}
Choose nonzero vectors $p_i\in\mathbb F^2$ representing $P_i$, and write $[u,v]=\det(u,v)$. Use the <cross-ratio> convention
$$
\boxed{(P_0P_1:P_2P_3)=\frac{[p_0,p_2][p_1,p_3]}{[p_0,p_3][p_1,p_2]}.}
$$
Distinctness makes the denominator nonzero. Every rescaling of a representative occurs once in numerator and denominator, so the value depends only on the <projective points>. Swapping both pairs simply exchanges the numerator's factors and the denominator's factors:
$$
(P_1P_0:P_3P_2)=\frac{[p_1,p_3][p_0,p_2]}{[p_1,p_2][p_0,p_3]}=(P_0P_1:P_2P_3).
$$
Under an invertible <linear map> $A$, $[Au,Av]=\det(A)[u,v]$. Two factors of $\det(A)$ occur above and below the fraction, proving <projective transformation> invariance.
Finally choose <homogeneous coordinates> making $P_0=[1:0]$, $P_1=[0:1]$, $P_2=[1:1]$. This normalization exists by the allowed theorem about a <projective frame>. A point distinct from $P_0$ is uniquely $[t:1]$, and the displayed <cross-ratio> is $t$. Thus equality of the two <cross-ratios> forces equal coordinates and
$$
\boxed{P_3=Q.}
$$
This also shows that for fixed first three distinct points the fourth point is determined by its <cross-ratio>.
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