The point minimising the sum of distances to the vertices of a convex quadrilateral

Lower bound. For every $P$, the triangle inequality gives
$$PA+PC\ge AC,\qquad PB+PD\ge BD.$$
Therefore $PA+PB+PC+PD\ge AC+BD$.

Equality. The first inequality is an equality iff $P$ lies on segment $AC$; the second iff $P$ lies on segment $BD$. Since the quadrilateral is convex, its diagonals intersect at a single point $X$ inside it. This point lies on both segments, so for $P=X$ both inequalities are equalities.

Answer. The minimum is $AC+BD$, attained exactly at the intersection $X$ of the diagonals. $\blacksquare$

Remark. For a non-convex quadrilateral, one of the vertices (the reflex one) gives the minimum instead.

Lesson. To minimise a sum of distances, pair the terms so that the triangle inequality gives a constant lower bound, then check when it is attained.