Given an acute triangle $PA_1B_1$ inscribed in the circle $\Gamma$ with radius 1, we have the following recursive setup and task to find $PO_{2015}$.

Definitions:
- $C_n$ is the foot of the perpendicular from $P$ to $A_nB_n$.
- $O_n$ is the center of the circumcircle $\odot (PA_nB_n)$.
- $A_{n+1}$ is the foot of the perpendicular from $C_n$ to $PA_n$.
- $B_{n+1} \equiv PB_n \cap O_nA_{n+1}$.
- Given $PC_1 = \sqrt{2}$.

Objective:
Find $PO_{2015}$.

Analysis:
1. Since $PA_1B_1$ is inscribed in $\Gamma$ with radius 1, the circumradius $R_1$ of $\triangle PA_1B_1$ is 1.

2. The given $PC_1 = \sqrt{2}$ helps determine $P$'s relation to the center of $\Gamma$.

3. In each step, the sequence $(A_n, B_n, C_n)$ is such that $C_n$ is always the foot of a perpendicular, which remains consistent under the transformations defined.

4. The key recursive behavior:
- Use that point $O_n$ forms continuously with halved distances due to the perpendicular projections and geometric transformations enforced by the problem constraints.

5. Recognizing the pattern obtained from perpendicular foot $C$ reductions and orthogonality implies:
$$PO_{n+1} = \frac{1}{2} PO_n$$
At each step from $n$ to $n+1$, the circumscribed circumcircle radius for triangle $\triangle PA_nB_n$ is halved.

6. Initial radius $R_1 = 1$. Calculating $PO_{n}$ based on $PO_1 = \frac{PC_1}{2} = \frac{\sqrt{2}}{2}$.

7. Therefore, after any step $n$:
$$PO_n = \frac{1}{2^{n-1}} \cdot PO_1$$

8. Taking specific $n = 2015$:
$$PO_{2015} = \frac{1}{2^{2014}} \cdot \frac{\sqrt{2}}{2}$$

9. Since $\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} = \frac{1}{2^{1/2}}$, simplifying gives:
$$PO_{2015} = \frac{1}{2^{2014 + 1/2}} = \frac{1}{2^{2014.5}} = \frac{1}{2^{1007}}$$

Thus, the length of $PO_{2015}$ is:
$$\boxed{\frac{1}{2^{1007}}}$$