My intuition as to why the statement is true is that the binomial distribution with the higher probability will always have more weight 'further right' if you were to plot it, however I can't turn this notion into a proof.
An equivalent statement is that $$\Pr[A = 0] + \cdots + \Pr[A = k - 1] \leq \Pr[B = 0] + \cdots + \Pr[B = k - 1].$$ All this is saying is that the binomial distribution with the lower probability as more weight 'further left'. Inductive arguments seem to go nowhere...
Any help would be much appreciated.
Let $q=rp$ for $r\in(0,1]$.
Let $A_1,A_2,\dots$ be Bernoulli-distributed with parameter $p$.
Let $U_1, U_2,\dots$ be Bernoulli-distributed with parameter $r$.
Also let all these rv's be independent.
Letting $B_i:=A_iU_i$ for $i=1,2,\dots$ observe that the $B_i$ are iid Bernoulli-distributed with parameter $rp=q$. And this with $B_i\leq A_i$.
Now let $A:=A_1+\cdots+A_n$ and let $B:=B_1+\cdots+B_n$.
Then $A\sim\operatorname{Bin}(n,p)$ and $B\sim\operatorname{Bin}(n,q)$.
But we have $B\leq A$ so that $\{B\geq k\}\subseteq\{A\geq k\}$ and consequently $\Pr(A\geq k)\geq\Pr(B\geq k)$.
Edit:
Troubles with the site at first prevented me from making the question a duplicate, so I decided to answer anyhow.