Lets assume we are given the four vector of momentum P which can be written as:
$$p = (p_0, p_t cos\theta , p_t sin \theta , P_L ) ........(1)$$
Where $P_L$ is the longitudinal competent. The transverse component can be written as easily : $p_t ^2 = p_t cos\theta^2 + p_t sin\theta ^2$
The differential form I need to show $$d^4p = 1/2 \ dp_0 \ dp_L \ d p_t^2 \ d \theta ........(2)$$
I understand the $p_L$ and $p_0$ are the two component that comes from $$d^4p = dp_0 \ dp_x \ dp_y dp_L .....(3)$$ Therefore $dp_x \ dp_y$ has to be equal to $d p_t^2 \ d \theta$ . How?
My intention is to get the equation (2) from equation(1) or (3).
Assuming you meant $p_t$ and not $q_t$, we have
$$\frac{1}{2}\: d(p_t^2)\:d\theta = p_t\:dp_t\:d\theta$$
which is just the polar coordinates Jacobian.