2. It is well known that $r \leq \frac{1}{2} R$. Therefore $\frac{1}{3}(1+r)^{2} \leq \frac{1}{3}\left(1+\frac{1}{2}\right)^{2}=\frac{3}{4}$. It remains only to show that $p \leq \frac{1}{4}$. We note that $p$ does not exceed one half of the circumradius of $\triangle A^{\prime} B^{\prime} C^{\prime}$. However, by the theorem on the nine-point circle, this circumradius is equal to $\frac{1}{2} R$, and the conclusion follows. Second solution. By a well-known relation we have $\cos A + \cos B + \cos C = 1 + \frac{r}{R} (=1 + r \text{ when } R=1)$. Next, recalling that the incenter of $\triangle A^{\prime} B^{\prime} C^{\prime}$ is at the orthocenter of $\triangle A B C$, we easily obtain $p = 2 \cos A \cos B \cos C$. Cosines of angles of a triangle satisfy the identity $\cos^{2} A + \cos^{2} B + \cos^{2} C + 2 \cos A \cos B \cos C = 1$ (the proof is straightforward: see (SL81-11)). Thus
$$\begin{aligned} p + \frac{1}{3}(1+r)^{2} & = 2 \cos A \cos B \cos C + \frac{1}{3}(\cos A + \cos B + \cos C)^{2} \\ & \leq 2 \cos A \cos B \cos C + \cos^{2} A + \cos^{2} B + \cos^{2} C = 1 \end{aligned}$$