Output power of a signal passing through an IIR filter
The analog frequency $\Omega=2\pi\times 623$ and the digital frequency is $\omega = \Omega T = \Omega/f_s = 0.03115\pi$
The frequency response at this frequency is
$$H(z)|_{z=e^{j\omega}} = H(e^{j0.03115\pi}) = \frac{0.2}{1-0.3 e^{-j0.03115\pi}}$$
and its square of magnitude is
$$
|H(e^{j0.03115\pi})|^2 = \Big|\frac{0.2}{1-0.3 e^{-j0.03115\pi}}\Big|^2 = 0.0812
$$
Now you'll see $P_y = P_x |H(e^{j0.03115\pi})|^2 = 0.0199$