Define an intersection point to be the intersection of two diagonals within $P$. If no three diagonals of $P$ are concurrent, then each intersection point $X$ is uniquely defined by the intersection of two diagonals $A_iA_k$ and $A_jA_l$, so $X$ is in the interior of quadrilateral $A_iA_jA_kA_l$. The diagonals of $P$ divide it into several regions. We wish to show that each intersection point $X=A_iA_k\cap A_jA_l$ may be assigned a unique region $R_X$ touching it. If this assignment is possible, then $R_X$ is contained in the quadrilateral $A_iA_jA_kA_l$, and we choose a point within $R_X$. The $\binom{n}{4}$ chosen points will all lie in different regions, so any segment connecting two of the chosen points must intersect some diagonal of $P$. Lemma. If some ($\geq 1$) intersection points are coloured blue, there is a region containing exactly one blue point on its perimeter. Proof of Lemma. Note that each intersection point touches exactly $4$ regions. Suppose each of the $4$ regions $R_1,R_2,R_3,R_4$ touching blue point $X$ have another blue vertex $X_1,X_2,X_3,X_4$ (all distinct from $X$, but there may be repeated points among them). If we extend the diagonals through $X$ to infinite lines, the entire plane is divided into four sectors, each containing $X_1,X_2,X_3,X_4$ respectively (they may lie on the boundaries of the respective sectors). Therefore, $X$ lies in the convex hull of $X_1X_2X_3X_4$. Thus, pick a blue point $X$ on the convex hull of the set of all blue points. It must touch a region that has no blue point other than $X$, which proves the Lemma. $\square$ Initially, colour all $\binom{n}{4}$ intersection points blue. Then, repeatedly apply the Lemma to find region $R_X$ with sole blue vertex $X$; assign $X$ to $R_X$, and remove the colour from $X$. Eventually, each intersection point is assigned a region touching it. If two intersection points $X$ and $Y$ were assigned the same region $R_X=R_Y$, where WLOG $X$ was assigned the region first, then $Y$ would have been a second blue vertex of $R_X$, contradiction. Therefore, the assigned regions are unique, and the question statement follows.