Let $E$ be the intersection point of the lines $AD$ and $BC$. As $|AD| = |BC|$, the triangle $ABE$ is isosceles with axis of symmetry the line through $X$ and $E$. In particular, $A$, $P$, $C$ are collinear iff $B$, $P$, $D$ are collinear.

According to the converse of Menelaus' Theorem for the triangle $EWY$, the three points $B$, $P$, $D$ are collinear iff
$$\frac{|ED|}{|DW|} \cdot \frac{|WP|}{|PY|} \cdot \frac{|BY|}{|BE|} = 1.$$

$$\frac{|CE|}{|CZ|} = \frac{|BE|}{|BX|}.$$

But this equation is true as the triangles $BEX$ and $CEZ$ are similar.

Note that $|AD| = |BC|$ implies that the line through $Z$ and $X$ is an axis of symmetry for the trapezium $ABCD$. In particular, $A$, $P$, $C$ are collinear iff $B$, $P$, $D$ are collinear.

Because the three lines $AB$, $WY$ and $DC$ are parallel, it follows that
$$\frac{|ZP|}{|PX|} = \frac{|CY|}{|YB|}.$$

As $|CY| = |CZ| = |DZ|$ and $|YB| = |BX|$, we obtain
$$\tan(\angle PBX) = \frac{|PX|}{|BX|} = \frac{|ZP|}{|DZ|} = \tan(\angle PDZ) = \tan(\angle DPW)$$
hence $\angle PBX = \angle DPW$, i.e. $B$, $P$, $D$ are collinear.