For polynomial lovers!

Clearly the other imaginary root is $r-si$ and third root is rational and let this third root be $'q'$ .

$P(x)=(x-q)(x-r-si)(x-r+si)=x^3-ax^2-bx-65$ (Here $i=\sqrt{-1}$ )

$\implies (x-q)(x^2-2rx+r^2+s^2)=x^3-ax^2-bx-65$

Compare coefficients to get $q(r^2+s^2)=65 \\ b=r^2+s^2+2rq\\ a=q+2r$

So each $p_{a,b}=a$ so we just need to find the values of $a$ in each pair.

If $q=1,r^2+s^2=65$ so $(r,s)=(\pm1,\pm8),(\pm8,\pm1),(\pm4,\pm7),(\pm7,\pm4)$

When $q=5, r^2+s^2=13$ so $(r,s)=(\pm2,\pm3),(\pm3,\pm2)$

When $q=13,r^2+s^2=5$ so in this case $(r,s)=(\pm2,\pm1),(\pm1,\pm2)$

Only $s^2$ appears in the equations for $a$ and $b$ so we ignore the signs of $s$ .

So the sum of the $p_{a,b}=a$ for $q=1$ is $q$ times the number of distinct $r$ values (as each value of $r$ generates a pair $(a,b)$ )

Answer= $(1)(8)+(5)(4)+(13)(4)=\boxed{80}$