Answer
$E_{release}=\left( M_{Parent}-M_{Daughter}-2m_e \right)c^2$.
Work Step by Step
The positron emission described is $^{A}_{Z}Parent \rightarrow\;^{A}_{Z-1}Daughter +\;^{0}_{1}e +\nu$.
We add Z electrons to the left hand side, the parent nuclear mass, so that we may use the atomic mass. To balance the equation, we add Z electrons to the right hand side.
We only need Z-1 electrons to be able to use the daughter’s atomic mass in our equation. We have 1 “extra” electron mass, and we also have the positron mass. In summary, on the right hand side, we have the daughter's atomic mass and 2 electron masses.
Calculate the energy released.
$$E_{release}=\left( M_{Parent}-(M_{Daughter}+2m_e) \right)c^2$$
$$E_{release}=\left( M_{Parent}-M_{Daughter}-2m_e \right)c^2$$
This was the relationship to be shown.
