How to show a binomial random variable dominates another binomial random variable with a smaller success value?

For $i=1,\dots n$ let $X_i$ have Bernouilli($p_h$) distribution.

For $i=1,\dots n$ let $U_i$ have Bernouilli($\frac{p_l}{p_h}$) distribution.

Also let it be that there is independency among these random variables.

For convenience let it be that $X_i(\omega),U_i(\omega)\in\{0,1\}$ for every $\omega\in \Omega$.

Now define $Y_i:=X_iU_i$ so that $Y_i\leq X_i$.

Then $Y_i$ has Bernouilli($p_l$) distribution.

If $X:=X_1+\cdots+X_n$ and $Y:=Y_1+\cdots+Y_n$ then they have the binomial distributions mentioned in your question.

Now observe that $X\geq Y$ so that $\{Y\geq\alpha\}\subseteq\{X\geq\alpha\}$ for any $\alpha$.

Consequently $\Pr(X\geq\alpha)\geq\Pr(Y\geq\alpha)$.

Let us consider the probability space $\Omega=[0,\,1]^n$ with $\sigma$-algebra of Borel subsets and Lebesgue measure as a probability measure.

Construct $2n$ Bernoulli r.v. $X_1,\ldots, X_n$ and $Y_1,\ldots, Y_n$ on this space such that they are independent within each vector, $\sum X_i \sim B(n, p_h)$, $\sum Y_i \sim B(n, p_l)$, and $X_i\geqslant Y_i$ for all $i=1,\ldots,n$.

Set for each $\vec \omega\in [0,\,1]^n$

$$X_i(\vec \omega)=\begin{cases}1, \ 0\leq \omega_i\leq p_h\cr 0, \ p_h < \omega_i\leq 1,\end{cases}$$ $$Y_i(\vec \omega)=\begin{cases}1, \ 0\leq \omega_i\leq p_l\cr 0, \ p_l < \omega_i\leq 1.\end{cases}$$

Then $Y_i\leq X_i$, $\sum Y_i \leq \sum X_i$. And therefore $$P(Y\geq a) = P\left(\sum Y_i\geq a\right) \leq P\left(\sum X_i\geq a\right)=P(X\geq a).$$

Other words, we construct each independent pairs $(X_i, Y_i)$ with Bernoulli distributions s.t. $\{Y_i=1\}$ imply $\{X_i=1\}$, but not vice versa.