= Solution
The variable initial <energy> $E_0$ replaces $m_\pi$ in the kinematic expression before differentiation. The charged-lepton <mass> remains fixed:
$$
p(E_0)=\frac{E_0^2-m_\ell^2}{2E_0}
=\frac{E_0}{2}-\frac{m_\ell^2}{2E_0}.
$$
Thus the required <Jacobian determinant> in this one-dimensional change of variables is
$$
\boxed{\left.\frac{dp}{dE_0}\right|_{E_0=m_\pi}
=\frac12+\frac{m_\ell^2}{2m_\pi^2}
=\frac{m_\pi^2+m_\ell^2}{2m_\pi^2}.}
$$
Equivalently, $E_0=\sqrt{p^2+m_\ell^2}+p$ gives $dE_0/dp=1+p/E_\ell$, so $dp/dE_0=E_\ell/(E_\ell+p)=E_\ell/m_\pi$, the same result. It is dimensionless in <natural units>.
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