3 mins to solve.. but we got there!

Lets consider p=1

If p=1, the sides are: 7,7,10
This triangle is valid per inequality (17>7, 14>10) so we don't have to worry about an invalid case.
In this case, x and y are the same angle. Their sum will therefore be 180-z (z is the third angle).
If this were a right isos triangle, the hypotenuse would be: \(7\sqrt{2}\), which is roughly 9.8. Since our actual hypotenuse is greater than 9.8, this implies that the angle z must be greater than 90. This, in turn, implies that x+y<90.
So for this case, we have QA<QB

Lets consider p=0
If p=0, the sides are: 6,8,10
-> valid triangle once again (14>10, 18>6, 16>8)
Notice that this is a 3:4:5 triangle -> this implies that the angle z is 90. Which in turn implies that x and y must sum to 90 in this case. Therefore, for this case, x+y=90 -> QA=QB.

As we have two valid cases, the answer is D.