jmc

Since the coefficients of $P(x)$ are real, if $r+si$ is a zero, then so is $r-si$. To avoid counting pairs of roots twice, we stipulate that $s>0$.

Letting $t$ denote the third root, we note that by Vieta's formulas, $$a=(r+si)+(r-si)+t=2r+t,$$so $t=a-2r$, which is an integer. By Vieta again, $$65=(r+si)(r-si)t=(r^2+s^2)t,$$so $r^2+s^2$ must be a positive divisor of $65$. Testing cases, we find that the possible values for $(r,s)$ are $(\pm 1,2)$, $(\pm 2,1)$, $(\pm 2,3)$, $(\pm 3,2)$, $(\pm 1,8)$, $(\pm 8,1)$, $(\pm 7,4)$, and $(\pm 4,7)$.

Now, given $r$ and $s$, we determine $p_{a,b}$. By Vieta's again, $$p_{a,b}=(r+si)+(r-si)+t=2r+t=2r+\frac{65}{r^2+s^2}.$$Over all possible pairs $(r,s)$, the $2r$ terms all cancel with each other. Looking at the list of possible pairs $(r,s)$, we get that the sum of all the $p_{a,b}$'s is $$4\left(\frac{65}{1^2+2^2}+\frac{65}{2^2+3^2}+\frac{65}{1^2+8^2}+\frac{65}{4^2+7^2}\right)=4(13+5+1+1)=\boxed{80}.$$