Ireland

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.

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Alternative solution.

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.