Notice that your equation is equivalent to
[;1 = \sum_{k=1}^{n-1} \left(\frac{k}{n}\right)^x = \sum_{k=1}^{n-1} \left(1-\frac{k}{n}\right)^x\right) ;]
We want to say that
[;\lim_{n \to \infty} \sum_{k=1}^{n-1} \left(1-\frac{k}{n}\right)^{n\log 2} = \sum_{k=1}^\infty \lim_{n \to \infty} \left(1-\frac{k}{n}\right)^{n\log 2} = \sum_{k=1}^\infty 2^{-k} = 1;]
To do this, consider the family of sequences ank defined by
[; a_n^k = \left(1-\frac{k}{n}\right)^{n\log 2} ;] for 0 < k < n and 0 otherwise.
Notice that since (1-k/n) and nlog 2, both considered as functions of n, is increasing for n > k, for fixed k a_nk is increasing as a function of n. Since
[;\lim_{n \to \infty} a_n^k = 2^{-k}, ;]
we have that ank <= 2-k for all n. Since this sequence is summable, the dominated convergence theorem lets us exchange the limits as we wanted to above, since of course
[;\sum_{k=1}^\infty a_n^k = \sum_{k=1}^{n-1} \left(1-\frac{k}{n}\right)^{n\log 2}.;]
In particular, for ε > 0 and n large,
[; \left|\sum_{k=1}^{n-1} \left(1-\frac{k}{n}\right)^{n\log 2} - 1\right| < \varepsilon;]
Plugging in the definition of P(n), we have equivalently that
[;\left|\sum_{k=1}^{n-1} \left(1-\frac{k}{n}\right)^{n\log 2} - \left(1-\frac{k}{n}\right)^{P(n)}\right| < \varepsilon;]
Fix a sufficiently large n, and assume that nlog 2 < P(n) (the other case is handled similarly and will give the same conclusion).
Then (1-k/n)P(n) <= (1-k/n)nlog2 for all k we care about. In particular, for k = 1,
[; 0 < \left(1-\frac{1}{n}\right)^{n\log 2} - \left(1-\frac{1}{n}\right)^{P(n)} < \varepsilon ;]
Rearranging, we get that
[;1-\frac{\varepsilon}{\left(1-\frac{k}{n}\right)^{n\log2}} < \left(1-\frac{k}{n}\right)^{P(n)-n\log 2};]
or
[;0 < P(n)-n\log 2 < \frac{\log\left(1-\frac{\varepsilon}{(1-1/n)^{n \log2}}\right)}{\log(1-1/n)};]
Since (1-1/n)nlog 2 stay bounded, the numerator of the right-hand side stays bounded away from 0, at least if we chose ε small enough (also (1-1/n)P(n) must stay bounded, in fact less than 1, so the other case is handled). The denominator goes to negative infinity. Since the numerator is negative, at least for large enough n, we have that P(n)-nlog2 goes to 0, and thus P(n)/n-log2 does, too.