Which of the conditions are necessary to determine the reason for the area? I have this statement:

My development was:
The $\triangle{BPC}$ have an area of $\frac{ah}{2}$, where $h =$ Height
The $\triangle{DPA}$ have an area of $\frac{at}{2}$, where $t = $ Height
The quadrilateral $ABCD$ have an area of $ak$, where $k =$ Height
So, area of $\triangle{BPC} + \triangle{DPA} = \frac{a}{2}(h+t)$
And the reason is $\frac{\frac{a(h+t)}{2}}{ak}$
This is only taking into account condition 1, but I could not achieve more.
The correct answer must be $D)$, but i cant understand how is it possible.
Then, ¿How can I get to the correct answer, and why?
 A: 
The quadrilateral $ABCD$ have an area of $ak$, where $k =$ Height

In the case of a parallelogram, the height is the distance between $\,BC\,$ and $\,DA\,$ so $\,k=h+t\,$.

And the reason is $\;\displaystyle\frac{\;\;\frac{a(h+t)}{2}\;\;}{ak}$

You must mean the ratio (not reason). Yes, which (again, for a parallelogram) simplifies to $\,\dfrac{1}{2}\,$.
A: You found this expression:
$$\frac{\left(\frac{a(h+t)}{2}\right)}{ak}.$$
You can convert this into a simpler expression that has just one horizontal bar
(a single fraction with a numerator and a denominator).
You should be able to simplify it further by cancellation.
To get a definite fraction of the area, you must also relate
$h + t$ to $k.$
Think about how the height of a triangle or parallelogram
(relative to a base $AD$ or $BC$) is measured,
and what that says about the relationship among $h,$ $t,$ and $k.$
To see this better, you may (for sake of argument)
assume either of the statements i) or ii);
then try  drawing 
the height of triangle $\triangle ADP$ on the base $AD,$
the height of triangle $\triangle BCP$ on the base $BC,$
and the height of parallelogram $ABCD$ between the bases $AD$ and $BC.$
