@chirudeepnamini @Shubhm
In $\Delta QPX \:\text{&}\:\Delta PYZ$,
$\frac{QP}{PY}=\frac{PX}{YZ}=\frac{1}{2}$
&, $\angle QPX=\angle PYZ$ {$\because$ corresponding angles of two parallel lines XP & ZY}
$\therefore$ $\Delta QPX \:\text{&}\:\Delta PYZ$ are similar (from $\text{SAS}$ similarity).
thus, angles of similar triangles are same.
So, $\angle YZP=\angle PXQ = 90^\circ$
Also, similar triangles have proportional sides,
so, $\frac{QP}{PY}=\frac{PX}{YZ}=\frac{XQ}{ZP}$
=>$\frac{XQ}{ZP}=\frac{1}{2}$
=> $ZP=8\:Km$.