Let $G^{\prime}$ be the centroid of triangle $A B C$. Recall the following. - Points $O, G^{\prime}, H$ lie on Euler's line of $\triangle A B C$ with $O G^{\prime}: G^{\prime} H=1: 2$. - Points $I, G^{\prime}, N$ lie on Nagel's line of $\triangle A B C$ with $I G^{\prime}: G^{\prime} N=1: 2$. Thus, $O I \parallel H N$ with $O I: H N=1: 2$. Therefore, we can detect parallel lines with ratio $2: 1$ in the figure. The only possible pairs are $P_{2} P_{4} \parallel P_{7} P_{5}$. Therefore, there are two possibilities: $\left(P_{2}, P_{7}\right)$ and $\left(P_{4}, P_{5}\right)$ must be $(O, H)$ and $(I, N)$ in some order. Intuitively, $H$ should be further out, so it's not unreasonable to guess that $P_{2}=O, P_{7}=H, P_{4}=I$, and $P_{5}=N$. Alternatively, perform the algorithm below with the other case to see if it fails. To identify the remaining points, we recall that the isogonal conjugate of $G$ and $N$ both lie on $O I$ (they are insimilicenter and exsimilicenter of incircle and circumcircle, respectively). Thus, $H, G, N, I$ lie on isogonal conjugate of $O I$, known as the Feuerbach's Hyperbola. It's also known that $O I$ is tangent to this line, and this hyperbola have perpendicular asymptotes. Using all information in the above paragraph, we can eyeball a rectangular hyperbola passing through $H, G, N, I$ and is tangent to $O I$. It's then not hard to see that $P_{6}=G$. Finally, we need to distinguish between symmedian and Kosnita points. To do that, recall that Kosnita point is isogonal conjugate of the nine-point center (not hard to show). Thus, $H, L, K, O$ lies on isogonal conjugate of $O H$, which is the Jerabek's Hyperbola. One can see that $H, L, K, O$ lies on the same branch. Moreover, they lie on this hyperbola in this order because the isogonal conjugates (in order) are $O$, centroid, nine-point center, and $H$, which lies on $O H$ in this order. Using this fact, we can identity $P_{5}=L$ and $P_{1}=K$, completing the identification.