= Solution
Take the dipoles to point along $+z$, and let their surface number density be $n_s$. An annulus of radius $\rho$ and width $d\rho$ contains $n_s2\pi\rho\,d\rho$ dipoles. The <electron>'s electrostatic <potential energy> is therefore
$$
U(z)=-\frac{en_sp}{4\pi\epsilon_0}\int_0^\infty
\frac{z\,2\pi\rho\,d\rho}{(\rho^2+z^2)^{3/2}}.
$$
For $z\ne0$, the integral without $2\pi$ equals $z/|z|$, giving
$$
U(z)=-\frac{en_sp}{2\epsilon_0}\operatorname{sgn}z.
$$
Thus the jump's magnitude is
$$
\boxed{|U(0^+)-U(0^-)|=\frac{en_sp}{\epsilon_0}
=\frac{e(p/A)}{\epsilon_0}.}
$$
For the chosen outward dipole orientation the electron-energy change from the negative to positive side is negative. The electrostatic voltage jump has the opposite sign. This distinguishes the requested magnitude from the sign of the <work function> change.
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