Answer: B

Solution:

Let $BC=x$ and $CD=AD=y$. Dropping the altitude from $A$ to $CD$ gives a right triangle with legs $x$ and $y-2$, and hypotenuse $y$. Therefore $$(y-2)^2+x^2=y^2,$$ so $x^2=4(y-1)$.

Since $x$ is an integer, write $x=2k$. Then $y=k^2+1$, and the perimeter is $$\begin{aligned} &p=2+x+y+y \\ &=2k^2+2k+4. \end{aligned}$$

We need $2k^2+2k+4<2015$, or $k^2+k<1005.5$. This holds for $k=1,2,\ldots,31$, while $k=32$ is too large. Thus there are $31$ possible perimeters.

Thus, B is the correct answer.