If the equilibrium constant for $A\rightleftharpoons B+C$ is ${\mathrm{K}}_{\mathrm{eq}}^{\left(A\right)}$ and that of $\mathrm{B}+\mathrm{C}\rightleftharpoons \mathrm{P}$is ${\mathrm{K}}_{\mathrm{eq}}^{\left(B\right)}$, the equilibrium constant for $\mathrm{A}\rightleftharpoons \mathrm{P}$ is :

$\mathrm{A}\rightleftharpoons \mathrm{B}+\mathrm{C} {\mathrm{K}}_{\mathrm{eq}}^{\left(1\right)}=\frac{\left[\mathrm{B}\right]\left[\mathrm{C}\right]}{\left[\mathrm{A}\right]}$

$\mathrm{B}+\mathrm{C}\rightleftharpoons \mathrm{P} {\mathrm{K}}_{\mathrm{eq}}^{\left(2\right)}=\frac{\left[\mathrm{P}\right]}{\left[\mathrm{B}\right]\left[\mathrm{C}\right]}$

For$\mathrm{A}\rightleftharpoons \mathrm{P} {\mathrm{K}}_{\mathrm{eq}}=\frac{\left[\mathrm{P}\right]}{\left[\mathrm{A}\right]}$

$\mathrm{A}\rightleftharpoons \mathrm{P} {\mathrm{K}}_{\mathrm{eq}}=\frac{\left[\mathrm{P}\right]}{\left[\mathrm{A}\right]}$

Multiplying equation (1) & (2)${\mathrm{K}}_{\mathrm{eq}}^{\left(A\right)}{\mathrm{K}}_{\mathrm{eq}}^{\left(B\right)}$

${\mathrm{K}}_{\mathrm{eq}}^{\left(1\right)}\times {\mathrm{K}}_{\mathrm{cq}}^{\left(2\right)}=\frac{\left[\mathrm{P}\right]}{\left[\mathrm{A}\right]}={\mathrm{K}}_{\mathrm{cq}}$

Hence the correct answer is (A)${\mathrm{K}}_{\mathrm{eq}}^{\left(A\right)}{\mathrm{K}}_{\mathrm{eq}}^{\left(B\right)}$.