Prove that $B^{\prime}, C^{\prime}, B, C$ are concyclic.
$\angle A C^{\prime} B^{\prime}=\angle A C B$.
Thus, $\triangle A C^{\prime} B^{\prime} \backsim \triangle A C B$.
Therefore, $\frac{B^{\prime} C^{\prime}}{B C}=\frac{A B^{\prime}}{A B}=\cos A$.
So, $B^{\prime} C^{\prime}=B C \cos A$
$=2 R \sin A \cos A$.
B
From $A^{\prime}, C^{\prime}, A, C$ being concyclic, we get $\angle B A^{\prime} C^{\prime}=\angle A$.
Similarly, $\angle B^{\prime} A^{\prime} C=\angle A$.
Thus, $\angle B^{\prime} A^{\prime} C^{\prime}=180^{\circ}-2 \angle A$.
Let the circumradius of $\triangle A^{\prime} B^{\prime} C^{\prime}$ be $R^{\prime}$. Then,
$$\frac{B^{\prime} C^{\prime}}{\sin \left(180^{\circ}-2 \angle A\right)}=2 R^{\prime},$$

which means $\frac{2 R \sin A \cos A}{\sin 2 A}=2 R^{\prime}$.
Hence, $R^{\prime}=\frac{1}{2} R=\frac{1}{2}$.
By Euler's inequality, $R^{\prime} \geqslant 2 p, p \leqslant \frac{R^{\prime}}{2}=\frac{1}{4}$.
By $R \geqslant 2 r, r \leqslant \frac{1}{2}$, we have
$\frac{1}{3}(1+r)^{2} \leqslant \frac{3}{4}$.
Thus, $1-\frac{1}{3}(1+r)^{2} \geqslant \frac{1}{4}$.
Therefore, $p \leqslant \frac{1}{4} \leqslant 1-\frac{1}{3}(1+r)^{2}$.
Hence, the original problem can be strengthened to $p \leqslant \frac{1}{4}$.