A point $M \in (AB)$ is uniquely determined by a real number $k \in (0, \infty)$ such that $\overline{AM} = k$, from which we get
$$\overrightarrow{OM} = \frac{1}{k+1}\overrightarrow{OA} + \frac{k}{k+1}\overrightarrow{OB}.$$
To find the points $N$ and $P$ uniquely, we need to find $x, y \in (0, \infty)$ such that $\overrightarrow{ON} = x, \overrightarrow{OP} = y$, and $O$ is the centroid of triangle $MNP$. From this, we have:
$$\overrightarrow{ON} = \frac{x}{x+1}\overrightarrow{OC} = -\frac{x}{x+1}\overrightarrow{OA} \quad \text{and} \quad \overrightarrow{OP} = \frac{y}{y+1}\overrightarrow{OD} = -\frac{y}{y+1}\overrightarrow{OB}.$$
Since $\overrightarrow{OA}$ and $\overrightarrow{OB}$ are not collinear, $O$ is the centroid of triangle $MNP$ if and only if $\frac{x}{x+1} = \frac{1}{k+1} \Leftrightarrow x = \frac{1}{k}$ and $\frac{y}{y+1} = \frac{k}{k+1} \Leftrightarrow y = k$, meaning that point $N$ is uniquely determined by the ratio $x = \frac{1}{k} = \frac{ON}{NC}$, and point $P$ is uniquely determined by the ratio $y = k = \frac{OP}{PD}$, which completes the problem.