Total binding energy (without considering repulsions),

$E_b=\left[Z{m}_p+\left(A-Z\right)m_n-m_x\right]c^2$

Where, ${}_Z{}^{A}X$ is the nuclei under consideration.

Now, considering repulsion :

Number of proton pairs $={}^{Z}C_2$

$\Rightarrow$ Thus repulsion energy $\propto \frac{Z\left(Z-1\right)}{2}\times \frac{1}{4\pi {ϵ}_0}\frac{e^2}{R}$

Where $R$ is the radius of the nucleus

$\Rightarrow E_b^p-E_b^n\propto Z\left(Z-1\right) \therefore$ there will be no repulsion term for neutrons.

Also, since $R=R_0{A}^{\frac{1}{3}}$

$\Rightarrow E_b^p-E_b^n\propto A^{-\frac{1}{3}}$

Because of repulsion among protons,

$E_b^p<E_b^n$

Since in ${\beta }^{+}$ decay, number of protons decrease $\Rightarrow$ repulsion would decrease

$\Rightarrow E_b^p$ increases