We can get the generating function for $P_n(x)$ and prove the property $P_n(x)=(-1)^{n+1}P_n(1-x)$ as follows.

Noticing that $\sum_p (-1)^{p+1} \frac{z^p}{p} = \log(1+z)$, we conclude that $$a_{n,k} = \sum_{p=1}^{n-k} \binom{n}{n-k-p} \frac{(-1)^{p+1}}{p} = [z^{n-k}]\, (1+z)^n\log(1+z).$$

Then, since $\binom{n+k}{n} = (-1)^k \binom{-n-1}{k}$, we get: \begin{split} P_n(x) &= [z^n]\, (1+xz)^{-n-1}(1+z)^n\log(1+z) \\ &=[z^n]\, \frac{\log(1+z)}{1+xz}\left(\frac{1+z}{1+xz}\right)^n. \end{split}

Using Lagrange inversion, we get: $$P_n(x) = [t^n]\ \frac{\log(1+\frac{t-1+\sqrt{(1-t)^2+4tx}}{2x})}{\sqrt{(1-t)^2+4tx}}.$$

It is now straightforward to verify that $P_n(1-x)=(-1)^{n+1}P_n(x)$.

We can get the generating function for $P_n(x)$ and prove the property $P_n(x)=(-1)^{n+1}P_n(1-x)$ as follows.

Noticing that $\sum_p (-1)^{p+1} \frac{z^p}{p} = \log(1+z)$, we conclude that $$a_{n,k} = \sum_{p=1}^{n-k} \binom{n}{n-k-p} \frac{(-1)^{p+1}}{p} = [z^{n-k}]\, (1+z)^n\log(1+z).$$

Then, since $\binom{n+k}{n} = (-1)^k \binom{-n-1}{k}$, we get: \begin{split} P_n(x) &= [z^n]\, (1+xz)^{-n-1}(1+z)^n\log(1+z) \\ &=[z^n]\, \frac{\log(1+z)}{1+xz}\left(\frac{1+z}{1+xz}\right)^n. \end{split}

Applying Lagrange-Burmann formula, we further obtain: $$P_n(x) = [t^n]\ \frac{\log(1+\frac{t-1+\sqrt{(1-t)^2+4tx}}{2x})}{\sqrt{(1-t)^2+4tx}}.$$

It is now straightforward to verify that $P_n(1-x)=(-1)^{n+1}P_n(x)$.