= Solution
Within the active set, cross-classify dosage $Z_i\in\{1,2\}$ against $Y_i$. Conditional on the vector $A=(A_1,\ldots,A_n)$, complete randomization assigns $n_1$ of the active patients to low dosage and $n_2$ to high dosage uniformly. Under independent assignment, active patients receive low and high dosage with conditional probabilities
$$
\frac{\pi_1}{\pi_1+\pi_2}
\quad\text{and}\quad
\frac{\pi_2}{\pi_1+\pi_2};
$$
conditioning further on their low- and high-dose totals again gives the same uniform allocation. Under $H_B$, active patients' outcomes are fixed as their common $Y_i(1)=Y_i(2)$, so the conditional Fisher p-value obeys
$$
\mathbb P(P_B\leq\alpha_2\mid A)\leq\alpha_2.
$$
Under $H_A$, $P_A$ is a function only of $A$ and the fixed outcomes. Applying the <law of iterated expectation> under the joint null gives
$$
\begin{aligned}
\mathbb P(P_A\leq\alpha_1,P_B\leq\alpha_2)
&=\mathbb E\!\left[
\mathbf1_{\{P_A\leq\alpha_1\}}
\mathbb P(P_B\leq\alpha_2\mid A)
\right]\\
&\leq\alpha_2\mathbb P(P_A\leq\alpha_1)
\leq\alpha_1\alpha_2.
\end{aligned}
$$
This conditional argument explains the stated near-independence even though the two tests reuse outcomes.
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