let angle QBA = x

therfore angle BAP = x (alternate angles in parallel lines...)

Similarly, let angle BQP = y

therfore angle APQ = y (alternate angles in parallel lines...)

Now, angle QTA = x + y ( the exterior triangle BQT is equal to the sum of the opposite interior angles)

Also, angle ATP = 180 - (x + y) ( angle sum of triangle PAT 180)

Therefore angle QTP = x + y + 180 - (x +y)

= 180

Thus Q,T,P are collinear