In the given equilibrium, the solid substances do not contribute to the equilibrium constant expression since their activities are considered to be 1.

For the reaction:

$ \text{A(s)} \rightleftharpoons \text{M(s)} + \frac{1}{2}\text{O}_2(\text{g}) $

The equilibrium constant, $ K_p $, is given by:

$ K_p = P_{\text{O}_2}^{n} $

where $ P_{\text{O}_2} $ is the partial pressure of $ \text{O}_2 $ and $ n $ represents the stoichiometric coefficient of $ \text{O}_2 $ in the balanced chemical reaction, which is $ \frac{1}{2} $. Therefore, the expression for $ K_p $ becomes:

$ K_p = \left( P_{\text{O}_2} \right)^{\frac{1}{2}} $

Given that $ K_p = 4 $, substituting into the equation gives:

$ 4 = \left( P_{\text{O}_2} \right)^{\frac{1}{2}} $

To find $ P_{\text{O}_2} $, square both sides of the equation:

$ 4^2 = P_{\text{O}_2} $

$ 16 = P_{\text{O}_2} $

Thus, the partial pressure of $ \text{O}_2 $ at equilibrium is $\boxed{16}$ atm.