WLOG assume $\angle BPC \ge \angle 120$ then we have $BC \ge \frac{\sqrt{3}}{2}(PB + PC)$ so $\frac{PB+PC}{BC} \le \frac{2}{\sqrt{3}}$. we also have $BC = (p-b) + (p-c)$ so $\frac{PB+PC}{(p-b) + (p-c)} \le \frac{2}{\sqrt{3}}$ and we know $\frac{PB+PC}{(p-b) + (p-c)} \ge \frac{PB}{p-b}$ or $\frac{PC}{p-c}$ so $ \min(\frac{PB}{p-b},\frac{PC}{p-c}) \le \frac{2}{\sqrt{3}}$. equality holds if $P$ is center of regular triangle $ABC$. we're Done.

If $P$ is the center of an equilateral triangle, then $\frac{PA}{p-a}=\frac{PB}{p-b}=\frac{PC}{p-c}=\frac{2}{\sqrt{3}}$. We'll show that this is the maximum possible value for $\min\left(\frac{PA}{p-a},\frac{PB}{p-b},\frac{PC}{p-c}\right)$. Assume the contrary and let there be a point $P\in\triangle ABC$ for which this value is $>\frac{2}{\sqrt{3}}$. Let $A_{1},B_{1},C_{1}$ be the touchpoints of the incircle of $\triangle ABC$ with sides $BC, CA, AB$ respectively. Then by Ptolemy's inequality for $AC_{1}PB_{1}$ we have that: $$PA\cdot B_{1}C_{1}\leq AB_{1}\cdot PC_{1}+AC_{1}\cdot PB_{1}=(p-a)(PB_{1}+PC_{1})<\frac{\sqrt{3}}{2}PA(PB_{1}+PC_{1})\Longrightarrow \frac{2}{\sqrt{3}}B_{1}C_{1}< (PB_{1}+PC_{1})\quad (1)$$Analogically, for $BA_{1}PC_{1}$ and $CB_{1}PA_{1}$ we derive: $$\frac{2}{\sqrt{3}}A_{1}C_{1}< (PA_{1}+PC_{1})\quad \text{and}\quad\frac{2}{\sqrt{3}}B_{1}C_{1}< (PA_{1}+PB_{1})$$Notice that $P$ is in the interior of $\triangle A_{1}B_{1}C_{1}$ as otherwise if WLOG $P\in \triangle AC_{1}B_{1}$, then $PA<AC_{1}=p-a\Longrightarrow \frac{PA}{p-a}<1<\frac{2}{\sqrt{3}}$, contradiction with the assumption. So we can say that $P\in \triangle A_{1}B_{1}C_{1}$. Now let's examine $(1)$ further and let $\angle B_{1}PC_{1}=\varphi$ now that we know that $P\in \triangle A_{1}B_{1}C_{1}$. Using the Cosine law for $\triangle B_{1}C_{1}P$ we have that: $$\frac{2}{\sqrt{3}}B_{1}C_{1}< (PB_{1}+PC_{1})\Longrightarrow \frac{4}{3}B_{1}C_{1}^2< (PB_{1}+PC_{1})^2$$$$\Longrightarrow \frac{1}{3}(PB_{1}^2+PC_{1}^2)<2PB_{1}PC_{1}\left(1+\frac{4}{3}\cos\varphi\right)\leq (PB_{1}^2+PC_{1}^2)\left(1+\frac{4}{3}\cos\varphi\right)$$$$\Longrightarrow \cos\varphi>\frac{1}{2}\Longrightarrow \varphi<120^{\circ}$$However, by manipulating the other two inequalities in the same way we get that $$\angle A_{1}PB_{1}+\angle B_{1}PC_{1}+\angle C_{1}PA_{1}<120^{\circ}+120^{\circ}+120^{\circ}=360^{\circ}$$, but $P\in \triangle A_{1}B_{1}C_{1}$, contradiction. Therefore: $$\min\left(\frac{PA}{p-a},\frac{PB}{p-b},\frac{PC}{p-c}\right)\leq \frac{2}{\sqrt{3}}$$