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Let $n:=N$. We have to estimate $$P(A_n>\sqrt p)=\sum_{j=k}^n a_j,$$ where $k:=k_{n,p}:=1+\lfloor n\sqrt p\rfloor>n\sqrt p$, $$a_j:=\binom nj p^jq^{n-j},$$ and $q:=1-p$. For $j\in\{k,\dots,n-1\}$, $$\frac{a_{j+1}}{a_j}=\frac{n-k}{k+1}\frac pq\le\frac{np}{n\sqrt p}\frac1q =\frac{\sqrt p}q\to0.$$ Therefore and by Stirling's formula, $$P(A_n>\sqrt p)\sim a_k\asymp\sqrt{\frac n{k(n-k)}}\exp\{-n D(\sqrt p||p)\},$$ where $$D(\sqrt p||p):=\sqrt p\,\ln\frac{\sqrt p}p+(1-\sqrt p)\ln\frac{1-\sqrt p}{1-p} \sim \sqrt p\,\ln\frac1{\sqrt p}.$$ Also, $$\sqrt{\frac n{k(n-k)}}\sim \sqrt{\frac1k}\asymp\exp\{o(n\sqrt p)\} =\exp\{o(n\sqrt p\,\ln\frac1{\sqrt p})\}.$$ Collecting pieces, we get $$P(A_n>\sqrt p)\asymp\exp\{-(1+o(1))n \sqrt p\ln\frac1{\sqrt p}\},$$ Thus, $$P(A_n>\sqrt p)\to0\iff n \sqrt p\,\ln\frac1{\sqrt p}\to\infty.$$