Kinetics of a simultaneous parallel radioactive decay
The question has already been solved by Yashwini and the answer given is correct.$^2$ A more intuitive and specific to question explanation would follow here.
Now, the two reactions given are:
\begin{array}{cc}
\require{cancel}
\ce{A -&gt; P} &amp;(t_{1/2} = 9\,\mathrm h) \\
\ce{A -&gt; Q} &amp;(t_{1/2} = 4.5\,\mathrm h) \\
\end{array}
Now using the rate law, we get,
\begin{align}
-\frac{\mathrm{d}[A]}{\mathrm{d}t}&amp;=k_\mathrm P [A] \tag{1} \\
-\frac{\mathrm{d}[A]}{\mathrm{d}t}&amp;=k_\mathrm Q [A] \tag{2} \\
\end{align}
The rate constant for a first order reaction having a half life of $t_{1/2}$ is defined as:
$$k=\frac{\ln 2}{t_{1/2}} \tag{3}$$
Now, substituting the given values of $t_{1/2}$ into the equations, we get $2k_\mathrm P = k_\mathrm Q$ (since $k\, \alpha \frac{1}{t_{1/2}})$
Now, intuitively since both reactions take place together, it would mean that for every one mole of P formed, two moles of Q forms. Therefore, for every mole of P formed, three moles of A react (since one mole is required for each mole of P and Q).
Now, we add the rate laws ($1$) and $(2)$, since the reactions take place simultaneously, to get:
$$-\frac{\mathrm{d}[A]}{\mathrm{d}t}=(k_\mathrm P +k_\mathrm Q) [A] \tag{4} $$
Now, since using the relation between $k_\mathrm{P}$ and $k_\mathrm{Q}$, we get $k_\mathrm{P} + k_\mathrm{Q} = 3k_\mathrm{P}$
Therefore using the integrated rate law for a first order reaction on equation $(4)$, we get:
$$A=A_0e^{-3k_\mathrm Pt} $$
Now, the amount of $A$ used here would be $A_0 -A$, and we get that value to be:
$$A_\text{used}=A_0\left(1-e^{-3k_\mathrm Pt}\right)$$
Now, as we have previously noted, for every three moles of A used, two moles  Q are formed. This means that the amount of Q now in the mixture would be two thirds of $A_\text{used}$. Therefore the amount of Q would be:
$$Q=\frac{2A_0\left(1-e^{-3k_\mathrm Pt}\right)}{3}$$
Now, we are given the condition, $Q = 2A$, substituting values of $Q$ and $A$ into the given relation we get:
$$\begin{align}
\frac{\cancel{2A_0}\left(1-e^{-3k_\mathrm Pt}\right)}{3} &amp;= \cancel{2A_0}\left(e^{-3k_\mathrm Pt}\right) \\
\implies 1 -e^{-3k_\mathrm Pt} &amp;= 3e^{-3k_\mathrm Pt} \\
\implies 4e^{-3k_\mathrm Pt} &amp;= 1
\end{align}$$
Solving for $t$, we get:
\begin{align}
3k_\mathrm Pt&amp;=2\ln 2 \\ \\
t&amp;=\frac{2\ln 2}{3k_\mathrm P}\\
\end{align}
Now, using equation $(3)$, we get the rate constant $k_\mathrm P$ to be $\frac{\ln 2}{9}$. Substituting this value into the expression for time, we get:
$$t=\frac{2 \cancel{\ln 2}}{\cancel{3} \frac{\cancel{\ln 2}}{\cancelto{3}{9}}}$$
Therefore, time taken for this condition to happen is:
$$t=2\times 3 = 6\ \mathrm h$$