$$ABCD$$ is a rectangle whose length $$AB$$ is $$20$$ units and breadth is $$10$$ units. Also, given $$AP = 8$$ units.

The area of the shaded region is $$\frac{p}{q}$$ sq unit, where $$p, q$$ are natural numbers with no common factors other than $$1$$. The value of $$p + q$$ is

We are given that $$AP=8$$ units and $$PB=12$$ units

Now, in triangles $$POB$$ and $$DOC$$, we have:

$$\angle BOP=\angle COD$$ [vertically opposite angles]

$$\angle BPO=\angle CDO$$ [alternate interior angle between the parallel lines $$AB$$ and $$CD$$]

$$\angle PBO=\angle OCD$$ [alternate interior angle between the parallel lines $$AB$$ and $$CD$$]

So, triangles $$POB$$ and $$DOC$$ are similar triangles

$$\therefore \dfrac{BO}{OC}=\dfrac{PB}{CD}=\dfrac{3}{5}$$

Let us assume that $$BO$$ is $$3x$$ units long

Then, $$OC$$ is $$5x$$ units long

Now, the area of triangle $$ABC=0.5*20*10=100$$ square units

Also, the area of triangle $$BOP$$ : area of triangle $$ABC=0.5*PB*OB*Sin(\angle ABC):0.5*AB*BC*Sin(\angle ABC)$$

The area of triangle $$BOP$$ : area of triangle $$ABC=0.5*12*3x*Sin(\angle ABC):0.5*20*8x*Sin(\angle ABC)$$

The area of triangle $$BOP$$ : area of triangle $$ABC=9:40$$

We know that area of triangle $$ABC$$ is $$100$$ square units

So, area of triangle $$BOP: 100=9:40$$

Hence, the area of triangle $$BOP$$ is $$=\dfrac{45}{2}$$ square units

And the area of quadrilateral $$BPOC=$$area of triangle $$ABC-$$area of triangle $$BOP$$

The area of quadrilateral $$BPOC=100-\dfrac{45}{2}=\dfrac{155}{2}$$ square units

Hence, $$p=155$$ and $$q=2$$ 

This makes $$p+q=155+2=157$$

Hence, Option C is correct