Real coefficients force the zeros to be $r \pm si$ and a real $q$ with $q(r^2 + s^2) = 65$

List the ways a factor of $65$ is a sum of two nonzero squares; for each, the choices $\pm r$ cancel in the sum, leaving only the $q$’s

Since $P$ has real coefficients, $r - si$ is also a zero, and the third zero $q$ is real. The sum of the zeros is $q + 2r = a,$ so $q = a - 2r$ is a nonzero integer. Their product is $q(r^2 + s^2) = 65,$ so $r^2 + s^2$ is a factor of $65.$ With $r$ and $s$ nonzero, the possibilities are $r^2 + s^2 = 5 = 1^2 + 2^2$ (with $q = 13$), $13 = 2^2 + 3^2$ (with $q = 5$), and $65 = 1^2 + 8^2 = 4^2 + 7^2$ (with $q = 1$).

For each representation $\{u, v\},$ the zero $r + si$ can have $r = \pm u$ or $\pm v,$ giving $4$ distinct polynomials (the sign of $s$ changes nothing). The sum of the zeros is $p_{a,b} = q + 2r,$ and over the four choices the $2r$ terms cancel, leaving $4q$ from each representation.

The total is $4 \cdot 13 + 4 \cdot 5 + 4 \cdot 1 + 4 \cdot 1 = 80.$