jmc

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 ${\left(\frac{4}{2}\right)}^2$ after the $-4x$ and ${\left(\frac{6}{2}\right)}^2$ after the $6y$. So we get $$4\left(x^2-4x+4\right)+16\left(y^2+6y+9\right)=-144+4\cdot 4+16\cdot 9=-144+16+144=16$$ Dividing both sides by $16$ gives $$\frac{{\left(x-2\right)}^2}{2^2}+\frac{{\left(y+3\right)}^2}{1^2}=1$$ This gives us $h=2$, $k=-3$, $a=2$, and $b=1$, so $h+k+a+b=\boxed{2}$