Answer: D

Small Hint:

Place $A=(0,0),$ $C=(15,0);$ then $B=\left(\tfrac{33}{5},\tfrac{56}{5}\right)$ and $P=(5,0)$

Big Hint:

One trapezoid needs $CD\parallel AB,$ the other needs $AE\parallel BC;$ find where each parallel line meets line $BP$

Solution:

Place $A=(0,0)$ and $C=(15,0).$ Then $B=\left(\tfrac{33}{5},\tfrac{56}{5}\right),$ and since $PC=10,$ $P=(5,0).$ Line $BP$ has slope $7,$ so it is $y=7(x-5).$

For $ABCD$ to be a trapezoid with $D$ on line $BP,$ take $CD\parallel AB.$ The line through $C$ parallel to $AB$ meets line $BP$ at $(\tfrac95,-\tfrac{112}{5}).$

For $ABCE$ with $E$ on line $BP,$ take $AE\parallel BC.$ The line through $A$ parallel to $BC$ meets line $BP$ at $(\tfrac{21}{5},-\tfrac{28}{5}).$

Their coordinate differences are $\tfrac{12}{5}$ and $\tfrac{84}{5},$ so $DE=\tfrac15\sqrt{12^2+84^2}=12\sqrt2.$

Thus, the correct answer is D.