Let $V$ indicate the volume, $S$ indicate area, $O$ be the center of its inscribed sphere, $H$ be the altitude,and $r$ be the desired radius. Then \begin{align} V_{P-ABC} &= V_{O-ABC} + V_{O-PAB} + V_{O-PBC} + V_{O-PCA}\nonumber\\ \frac{1}{3}\cdot H\cdot S_{ABC} &= \frac{1}{3}\cdot r\cdot (S_{ABC} + S_{PAB} + V_{PBC} + V_{PCA}) \end{align} From the given conditions: $$AB=BC=CA=1 \implies S_{ABC}=\frac{\sqrt{3}}{4}$$ \begin{align*} PM^2 &= PO^2 + OM^2 \\ &= PO^2 + \Big(\frac{1}{3}\cdot AM\Big)^2\\ &= (\sqrt{2})^2 + \Big(\frac{1}{3}\cdot \frac{\sqrt{3}}{2}\Big)^2\\ &=\frac{25}{12} \end{align*} $$S_{PBC}=\frac{1}{2}\cdot BC\cdot PM = \frac{1}{2}\cdot 1 \cdot \frac{5}{2\sqrt{3}}=\frac{5}{4\sqrt{3}}$$ Then, $$\sqrt{2}\cdot \frac{\sqrt{3}}{4} = r\cdot \Big(\frac{\sqrt{3}}{4} + 3\cdot \frac{5}{4\sqrt{3}}\Big)\implies r=\boxed{\frac{\sqrt{2}}{6}}$$