First, group the terms as follows: $$\left(4x^2-16x\right)+\left(16y^2+96y\right)=-144$$ Factoring out the coefficients of $ x^2 $ and $ y^2 $ gives $$4\left(x^2-4x\right)+16\left(y^2+6y\right)=-144$$ To complete the square, we need to add $ \left(\dfrac{4}{2}\right)^2 $ after the $ -4x $ and $ \left(\dfrac{6}{2}\right)^2 $ after the $ 6y $. So we get $$4\left(x^2-4x+4\right)+16\left(y^2+6y+9\right)=-144+4\cdot4+16\cdot9=-144+16+144=16$$ Dividing both sides by $ 16 $ gives $${\dfrac{\left(x-2\right)^2}{2^2}+\dfrac{\left(y+3\right)^2}{1^2}=1}$$ This gives us $ h=2 $, $ k = -3 $, $ a=2 $, and $ b=1 $, so $ h+k+a+b = \boxed{2} $