... my question is why can we use <PERSON>'s rotational second law in $P$'s frame? The point $P$ of contact moves (indeed, accelerates, with time). I suppose that, at an instant of time, we can consider $P$ as the point on the fixed ground which is in contact with the disk.
When in doubt, be careful; assumptions here do not appear as they seem.
Let $\textbf{r}{\text{s}}$ be the vector connecting our lab frame to some other stationary frame of reference, and let all positions in the lab frame $\textbf{r}_i$ be represented as $\textbf{r}\text{s} + \textbf{δ}i$, where $\textbf{δ}_i = \textbf{r}_i - \textbf{r}\text{s}$.
Angular momentum of a body, in that other frame of reference, is defined by $$\textbf{L} = \int_{\,V}\textbf{δ}{dm} \times (\dot{\textbf{δ}}{dm}\, dm) = \int_{\,V}(\textbf{r}{dm} - \textbf{r}\text{s})\times\dot{\textbf{r}}_{dm}\,dm$$
With that in mind, let's examine the validity of switching stationary frames. 101
For constant different $\textbf{r}_\text{s}$, it might be of value to find the geometric shape such that $\textbf{L}$ remains invariant.
Then the angular momentum, for different $\textbf{r}\text{s}$, becomes $$\begin{align} \textbf{L} &= \int{\,V}(\textbf{r}{dm}\times\dot{\textbf{r}}{dm})\,dm - \int_{\,V}(\textbf{r}\text{s}\times\dot{\textbf{r}}{dm})\,dm \ &= \int_{\,V}(\textbf{r}{dm}\times\dot{\textbf{r}}{dm})\,dm - \textbf{r}\text{s}\times\left(\int{\,V}dm\right)\left(\frac{\int_{\,V}\dot{\textbf{r}}{dm}\,dm}{\int{\,V}dm}\right) \ &= \int_{\,V}(\textbf{r}{dm}\times\dot{\textbf{r}}{dm})\,dm - \textbf{r}\text{s}\times M\dot{\textbf{r}}\text{CM} \ \end{align}$$
Due to the cross product, the endpoints of $\textbf{r}\text{s}$ that correspond to the same $\textbf{L}\,$ effectively draws a line parallel to whichever direction $\dot{\textbf{r}}\text{CM}$ points in.
Hence, if we were to consider a disk rolling down a slope, we can set the endpoints of $\textbf{r}\text{s}$ to sit on the contact points the disk rolls through, since this is a straight line. As the disk rolls by each endpoint, we can peer in each stationary frame and safely analyze the torques and forces as expected, in a snapshot of time; the contributions $\dot{\textbf{L}}=\textbf{Γ}$ considered correspond to a commensurate change in the same angular momentum across our stationary $\textbf{r}\text{s}$ vectors.
$$\rule[0]{300pt}{0.4pt}$$
It's important to note that if the endpoint of $\textbf{r}s$ is legitimately translating along with the contact point of the disk, then the angular momentum obtained is different from the angular momentum we analyzed through jumping across stationary frames—we must take into account $\dot{\textbf{r}}\text{s} \neq 0$, and the equivalence between $\dot{\textbf{L}}$ and $\textbf{Γ}$ is not quite true.
... the result holds only in an inertial frame precisely because <PERSON>'s second law has been invoked (with the standard "coincidence"/exception of the result holding in the CM frame being noted). 804