As seen in Fig. 4.1, suppose the projections of the inscribed sphere's center $O$ on faces $ABC$ and $ABP$ are $H$, $K$, respectively, the midpoint of $AB$ is $M$, and the radius of the sphere is $r$. Then $P$, $K$, $M$ are collinear, $\angle PHM = \angle PKO = \frac{\pi}{2}$, and

Fig. 4.1
$$OH = OK = r, \quad PO = PH - OH = \sqrt{2} - r, \\ MH = \frac{\sqrt{3}}{6} AB = \frac{\sqrt{3}}{6}, \quad PM = \sqrt{MH^2 + PH^2} = \sqrt{\frac{1}{12} + 2} = \frac{5\sqrt{3}}{6}.$$
Then we have
$$\frac{r}{\sqrt{2} - r} = \frac{OK}{PO} = \sin \angle KPO = \frac{MH}{PM} = \frac{1}{5}.$$
Therefore, $r = \frac{\sqrt{2}}{6}$.
The answer is $\frac{\sqrt{2}}{6}$.