Answer: B

Solution:

In every such quadrilateral $CD \ge AB.$ Let $E$ be the foot of the perpendicular from $A$ to $\overline{CD};$ then $CE = 2$ and $AE = BC.$ Let $x = AE$ and $y = DE,$ so $AD = 2 + y.$

By the Pythagorean Theorem $x^2 + y^2 = (2+y)^2,$ so $x^2 = 4 + 4y$ and $x$ is even. Writing $x = 2z$ gives $y = z^2 - 1,$ and the perimeter is $$x + 2y + 6 = 2z^2 + 2z + 4.$$

Increasing values $z = 1, 2, 3, \dots$ give the required quadrilaterals with increasing perimeter. For $z = 31$ the perimeter is $1988,$ and for $z = 32$ it is $2116.$ Therefore there are $31$ possible values of $p \lt 2015.$

Thus, the correct answer is B.