Finding the inverse Inverse $\mathcal Z$-transform
Multiply the top and bottom of the fraction by $z^{-60k}$ in each term, and you are left with each term being:

$$ z^{-60k}\frac{1}{z-a} = z^{-60k-1}\frac{1}{1-az^{-1}}
$$

The first factor delays the output by $60k +1$ samples, and the remaining factor is a simple one pole transfer function, giving:

$$  x[n] = \sum_{k=1}^{P} a^{n - 60k - 1} u[n - 60k - 1]
$$