Answer
$(1-pq)e^{p+q}$
Work Step by Step
Since, $Jacobian =\begin{vmatrix} \dfrac{\partial x}{\partial p}&\dfrac{\partial x}{\partial q}\\\dfrac{\partial y}{\partial p}&\dfrac{\partial y}{\partial q}\end{vmatrix}$
Given that $x=pe^q$
Thus, $\dfrac{\partial x}{\partial p}=e^q$ and $\dfrac{\partial x}{\partial q}=p e^q$
and Given: $y=qe^p$
Thus, $\dfrac{\partial y}{\partial p}=q e^p$ and $\dfrac{\partial y}{\partial q}= e^p$
Now, $Jacobian =\begin{vmatrix} \dfrac{\partial x}{\partial p}&\dfrac{\partial x}{\partial q}\\\dfrac{\partial y}{\partial p}&\dfrac{\partial y}{\partial q}\end{vmatrix}=\begin{vmatrix} e^q&pe^q\\ qe^p&e^p\end{vmatrix}$
or, $=e^p \times e^q-qe^p \times pe^q$
or, $=e^{p+q}-pqe^{p+q}$
or, $=(1-pq)e^{p+q}$
