Find him guilty

As in the problem, let $N=2k+1.$ Let $p = p_N$ just for ease of notation. The probability that the criminal is found guilty is $\sum_{i=k+1}^{2k} \binom{N}{i} p^i (1-p)^{N-i}.$ Let's take the derivative of this with respect to $p$ and set it equal to $0.$ This gives $0 = \sum_{i=k+1}^{2k} \binom{N}{i} \left(ip^{i-1}(1-p)^{N-i} - (N-i)p^i (1-p)^{N-i-1}\right).$ Break this into two sums: $0 = \sum_{i=k+1}^{2k} i \binom{N}{i} p^{i-1}(1-p)^{N-i} - \sum_{i=k+1}^{2k} (N-i) \binom{N}{i} p^i (1-p)^{N-i-1}.$ Change the index on the first one: $0 = \sum_{j=k}^{2k-1} (j+1)\binom{N}{j+1} p^j(1-p)^{N-j-1} - \sum_{i=k+1}^{2k} (N-i) \binom{N}{i} p^i (1-p)^{N-i-1}.$ Now change the $j$ back to an $i$ and recombine, remembering to take out the terms at the edge: $0 = (k+1)\binom{N}{k+1} p^k(1-p)^k - N p^{2k} + \sum_{i=k+1}^{2k-1} \left( (i+1) \binom{N}{i+1} - (N-i)\binom{N}{i} \right) p^i (1-p)^{N-i-1}.$ Magically, the quantity $(i+1)\binom{N}{i+1} - (N-i)\binom{N}{i}$ equals $\frac{N!}{i!(N-i-1)!} - \frac{N!}{i!(N-i-1)!} = 0,$ so that whole sum on the right goes away. We're left with $\begin{aligned} Np^{2k} &= (k+1)\binom{N}{k+1} p^k(1-p)^k \\ \left( \frac{p}{1-p} \right)^k &= \frac{k+1}{N} \binom{N}{k+1} \\ \left( \frac{p}{1-p} \right)^k &= \binom{N-1}{k} = \binom{2k}{k} \\ \frac{p}{1-p} &= \binom{2k}{k}^{1/k}. \end{aligned}$

Now we're ready to use Stirling's approximation applied to central binomial coefficients , which says that $\binom{2k}{k} \sim \frac{4^k}{\sqrt{\pi k}}.$ So $\frac{p}{1-p} \sim \frac4{(\pi k)^{1/2k}},$ which approaches $4$ as $k \to \infty.$ Then $\frac{p}{1-p} \to 4$ gives $p \to \fbox{4/5}.$

(Bonus: you might wonder if there is a natural explanation for the magic cancellation above, where the middle terms of the two sums are the same. Try to find it!)