We have two reactions:

$ A \;\rightleftharpoons\; B + C \quad\text{with } K_{eq}^{(1)}$

$ B + C \;\rightleftharpoons\; P \quad\text{with } K_{eq}^{(2)}$

We want the equilibrium constant for the overall reaction

$ A \;\rightleftharpoons\; P. $

How to combine the equilibrium constants

When two reactions are added to yield a net (overall) reaction, the equilibrium constant of the net reaction is the product of the equilibrium constants of the individual reactions. Formally:

$ \text{(Reaction 1)} + \text{(Reaction 2)} \;\Rightarrow\; \text{(Net Reaction)}, $

and

$ K_{\text{net}} \;=\; K_{\text{(Reaction 1)}} \times K_{\text{(Reaction 2)}}. $

Applying it here

Reaction 1: $A \to B + C$ has $K_{eq}^{(1)}.$

Reaction 2: $B + C \to P$ has $K_{eq}^{(2)}.$

When we add them,

$ \underbrace{A}_{\text{Reactant 1}} \;\longrightarrow\; \underbrace{B + C}_{\text{Products of Reaction 1}=\text{Reactants of Reaction 2}} \;\longrightarrow\; \underbrace{P}_{\text{Product of Reaction 2}}. $

So the net reaction is $A \to P$, and its equilibrium constant is

$ K_{eq}^{\text{(net)}} \;=\; K_{eq}^{(1)} \times K_{eq}^{(2)}. $

Thus, the correct answer is:

$ \boxed{K_{eq}^{(1)} \, K_{eq}^{(2)}}. $

Hence, the answer (Option D) is correct.