Rewrite the right hand side as $$\frac1{2^s-1}=\left(\frac12\right)^s+\left(\frac14\right)^s+\left(\frac18\right)^s+\ldots.$$ Then your inequality reads as $$\sum_{i=1}^p f(2^{n_i}/N)\leqslant \sum_{i>0} f(1/2^i)\quad\quad\quad(\heartsuit)$$ for a function $f(t)=t^s$. I claim that $(\heartsuit)$ holds for every concave function $f$. In other words, the set $\{2^{n_i}/N\colon i=1,2,\ldots,p\}$ majorizes the (infinite) set $\{1/2,1/4,\ldots\}$. This is clear: for every $k=1,2,\ldots,p$ we have $$\sum_{i=1}^k\frac{2^{n_i}}N=\frac{2^{n_1}+\ldots+2^{n_k}}{N}=\left(1+\frac{2^{n_{k+1}}+\ldots+2^{n_p}}{2^{n_{1}}+\ldots+2^{n_{k}}}\right)^{-1}\geqslant \left(1+\frac{2^{n_{k}}-1}{2^{n_{k}}(2^k-1)}\right)^{-1}\\ >\left(1+\frac{1}{2^k-1}\right)^{-1}=1-\frac1{2^k}=\frac12+\frac14+\ldots+\frac1{2^k}.$$

Rewrite the right hand side as $$\frac1{2^s-1}=\left(\frac12\right)^s+\left(\frac14\right)^s+\left(\frac18\right)^s+\ldots.$$ Then your inequality reads as $$\sum_{i=1}^p f(2^{n_i}/N)\leqslant \sum_{i>0} f(1/2^i)\quad\quad\quad(\heartsuit)$$ for a function $f(t)=t^s$. I claim that $(\heartsuit)$ holds for every concave function $f$ with $f(0)=0$. In other words, the infinite multiset $\{2^{n_i}/N\colon i=1,2,\ldots,p;0,0,\ldots\}$ majorizes the set $\{1/2,1/4,\ldots\}$. This is clear: for every $k=1,2,\ldots,p$ we have $$\sum_{i=1}^k\frac{2^{n_i}}N=\frac{2^{n_1}+\ldots+2^{n_k}}{N}=\left(1+\frac{2^{n_{k+1}}+\ldots+2^{n_p}}{2^{n_{1}}+\ldots+2^{n_{k}}}\right)^{-1}\geqslant \left(1+\frac{2^{n_{k}}-1}{2^{n_{k}}(2^k-1)}\right)^{-1}\\ >\left(1+\frac{1}{2^k-1}\right)^{-1}=1-\frac1{2^k}=\frac12+\frac14+\ldots+\frac1{2^k}.$$ The infinite version of Karamata inequality is proved by the same Abel transform trick as the proof given in Wikipedia.