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 \[4 \left(\frac{65}{1^2+2^2} + \frac{65}{2^2+3^2} + \frac{65}{1^2+8^2} + \frac{65}{4^2+7^2}\right) = 4 (13 + 5 + 1 + 1) = \boxed{80}.\]