Fatima traverses like this:-

Making right angle triangles, $\unicode{0x25FA} \: PXQ$ & then $ \unicode{0x25FA} \: PZY$.
So, to reach point $P$ from $Y$ she has to travel =$\sqrt{(10^2-6^2)} = 8 \: km$. (in East direction)
Hence, Correct Answer: $ A. \: 8 \: km, \: East $
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, $\frac{QP}{PY}=\frac{PX}{YZ}=\frac{XQ}{ZP}$
=>$\frac{XQ}{ZP}=\frac{1}{2}$
=> $ZP=8\:Km$.