If the equilibrium constant for $$A \rightleftharpoons B + C$$ is $$K_{eq}^{(1)}$$ and that of $$B + C \rightleftharpoons P$$ is $$K_{eq}^{(2)}$$, the equilibrium constant for $$A \rightleftharpoons P$$ is:

For the first equilibrium $$A \;\rightleftharpoons\; B + C$$ we recall the definition of the equilibrium constant for a reaction of the type $$\alpha\,A + \beta\,B \;\rightleftharpoons\; \gamma\,C + \delta\,D$$, namely $$K_{eq} = \dfrac{[C]^{\gamma}[D]^{\delta}}{[A]^{\alpha}[B]^{\beta}}.$$

Applying this definition to the reaction $$A \;\rightleftharpoons\; B + C$$, we have

$$K_{eq}^{(1)} = \dfrac{[B][C]}{[A]}.$$

From this, we can rearrange to express the product concentration $$[B][C]$$ in terms of $$[A]$$:

$$[B][C] = K_{eq}^{(1)}[A].$$

Now we consider the second equilibrium $$B + C \;\rightleftharpoons\; P$$. Using the same definition of an equilibrium constant, we write

$$K_{eq}^{(2)} = \dfrac{[P]}{[B][C]}.$$

We wish to eliminate $$[B][C]$$ so that only $$[A]$$ and $$[P]$$ remain. Substituting the expression $$[B][C] = K_{eq}^{(1)}[A]$$ obtained from the first equilibrium into the equation for $$K_{eq}^{(2)}$$ gives

$$K_{eq}^{(2)} = \dfrac{[P]}{K_{eq}^{(1)}[A]}.$$

Now we rearrange this equation to isolate the ratio $$\dfrac{[P]}{[A]}$$, because that ratio is precisely the equilibrium constant we seek for the overall process $$A \;\rightleftharpoons\; P$$:

Multiplying both sides by $$K_{eq}^{(1)}[A]$$, we obtain

$$K_{eq}^{(2)} K_{eq}^{(1)} [A] = [P].$$

Dividing both sides by $$[A]$$, we get

$$\dfrac{[P]}{[A]} = K_{eq}^{(1)} K_{eq}^{(2)}.$$

Again using the definition of the equilibrium constant for the simple reaction $$A \;\rightleftharpoons\; P$$, we know that

$$K_{eq\,(A \rightleftharpoons P)} = \dfrac{[P]}{[A]}.$$

Therefore, we can directly identify

$$K_{eq\,(A \rightleftharpoons P)} = K_{eq}^{(1)} K_{eq}^{(2)}.$$

Hence, the correct answer is Option D.