Michaelis-Menten rate law for enzyme which catalyzes two reactions: steady state?
In the steady-state reaction, the intermediate concentration [ES] is assumed to remain at a small constant value. So in this case only if k2 >> k1 and similar for the second reaction. ES is now a reactive intermediate and there is no stable equilibrium between S, E and P.



\begin{align}
\frac{d[\ce{S1E}]}{dt} &amp;= \ce{k1}[\ce{S1}][\ce{E}] - k_{-1} [\ce{S1E}] - \ce{k2}[\ce{S1E}] = 0\\
\end{align}

Therfore:

\begin{align}
[\ce{S1E}] &amp;= \frac{\ce{k1}}{k_{-1}+\ce{k2}}[\ce{S1}][\ce{E}] = K_a[\ce{S1}][\ce{E}]
\\
\end{align}

Similarly for the second reaction:

\begin{align}
[\ce{S2E}] &amp;= \frac{\ce{k3}}{k_{-3}+\ce{k4}}[\ce{S2}][\ce{E}] = K_b[\ce{S2}][\ce{E}]
\\
\end{align}

The enzyme E is involved in both reactions and its total concentration(bound and unbound) is constant. This total concentration of enzyme $[E]_{0}$ is equivalent to the concentration of the free enzyme before adding the substrates. The concentration of the free enzyme at a certain time t is [E]:

\begin{align}
[E]_{0} &amp;= [E] + [\ce{S1E}] + [\ce{S2E}]
\\
\end{align}

if you substitute $[\ce{S1E}]$ and $[\ce{S2E}]$ with the previous expressions then:

\begin{align}
[E] &amp;= \frac{[E]_{0}}{1 + k_{a}[\ce{S1}]+k_{b}[\ce{S2}]}
\\
\end{align}

Eventually:
\begin{align}
\frac{d[P_{1}]}{dt} = k_2[\ce{S1E}] = \ce{k2}(K_a[\ce{S1}][\ce{E}]) = K_p[\ce{S1}][\ce{E}] = K_p[\ce{S1}]\frac{[E]_{0}}{1 + k_{a}[\ce{S1}]+k_{b}[\ce{S2}]}
\\
\end{align}

and similarly for the rate of production of $P_2$