Solution:

Observe that the value $$AE = AB-EB$$$$= 5-1$$$$=4.$$ Also, the value $$FC = DC-DF$$$$= 5-2$$$$=3.$$ Then, since $AE \mid \mid FC,$ we know $\triangle AEP \sim \triangle CFP$ by angle angle symmetry. Thus, $$\dfrac{PF}{PE} = \dfrac{FC}{EA} = \dfrac 34 .$$ This makes $$\dfrac{PF}{EF} = \dfrac{3}{3+4} = \dfrac 37.$$

Then, let $X$ be the extension of $AG$ to meet $CD.$ Since two sides are parallel, we have $$\dfrac{AD}{DX} = \dfrac{BG}{AB}.$$ Thus, $$\dfrac{4}{DX} = \dfrac{3}{5}$$ $$DX = \dfrac{20}3.$$ This makes $$FX = \dfrac{20}3 - 2 = \dfrac{14}3 .$$ Then, since $AE \mid \mid FX,$ we know $\triangle AEQ \sim \triangle XFQ$ by angle angle symmetry. Therefore, $$\dfrac{QF}{QE} = \dfrac{FX}{EA} = \dfrac {\frac{14}{3}}4 = \dfrac 76.$$ This makes $$\dfrac{QF}{EF} = \dfrac{7}{7+6} = \dfrac 7{13}.$$

As such, $$\dfrac{QP}{EF} = \dfrac{QF}{EF} - \dfrac{PF}{EF}$$$$= \dfrac{7}{13} - \frac 37$$$$= \dfrac{10}{91}.$$

Thus, the correct answer is D .