On the plane $S$ in a space, given are unit circle $C$ with radius 1 and the line $L$. Find the volume of the solid bounded by the curved surface formed by the point $P$ satifying the following condition $(a),\ (b)$.

$(a)$ The point of intersection $Q$ of the line passing through $P$ and perpendicular to $S$ are on the perimeter or the inside of $C$.

$(b)$ If $A,\ B$ are the points of intersection of the line passing through $Q$ and pararell to $L$, then $\overline{PQ}=\overline{AQ}\cdot \overline{BQ}$.
1. **Clarify the problem statement**: We need to find the volume of the solid bounded by the curved surface formed by the point \( P \) satisfying the given conditions. The conditions are:
   - (a) The point of intersection \( Q \) of the line passing through \( P \) and perpendicular to \( S \) is on the perimeter or inside the unit circle \( C \).
   - (b) If \( A \) and \( B \) are the points of intersection of the line passing through \( Q \) and parallel to \( L \), then \(\overline{PQ} = \overline{AQ} \cdot \overline{BQ}\).

2. **Orient the plane**: Let the plane \( S \) be the \( xy \)-plane, the unit circle \( C \) be given by \( x^2 + y^2 = 1 \), and the line \( L \) be parallel to the \( x \)-axis.

3. **Identify points \( Q \), \( A \), and \( B \)**:
   - Let \( Q = (x, y) \) be a point on or inside the unit circle.
   - Since \( L \) is parallel to the \( x \)-axis, the points \( A \) and \( B \) are the intersections of the line through \( Q \) parallel to \( L \) with the circle \( C \). Thus, \( A = (-\sqrt{1-y^2}, y) \) and \( B = (\sqrt{1-y^2}, y) \).

4. **Calculate \(\overline{PQ}\)**:
   - Given \(\overline{PQ} = \overline{AQ} \cdot \overline{BQ}\), we have:
     \[
     \overline{AQ} = x + \sqrt{1-y^2}, \quad \overline{BQ} = \sqrt{1-y^2} - x
     \]
     \[
     \overline{PQ} = (x + \sqrt{1-y^2})(\sqrt{1-y^2} - x) = 1 - x^2 - y^2
     \]

5. **Determine the coordinates of \( P \)**:
   - Since \(\overline{PQ} = 1 - x^2 - y^2\), the point \( P \) can be either above or below the plane \( S \):
     \[
     P = (x, y, 1 - x^2 - y^2) \quad \text{or} \quad P = (x, y, -(1 - x^2 - y^2))
     \]

6. **Calculate the volume of the solid**:
   - The volume of the solid is given by integrating the height \( 1 - x^2 - y^2 \) over the area of the unit circle, and then doubling it to account for both above and below the plane:
     \[
     V = 2 \iint_{x^2 + y^2 \le 1} (1 - x^2 - y^2) \, dx \, dy
     \]

7. **Convert to polar coordinates**:
   - Using polar coordinates \( x = r \cos \theta \) and \( y = r \sin \theta \), the integral becomes:
     \[
     V = 2 \int_{0}^{2\pi} \int_{0}^{1} (1 - r^2) r \, dr \, d\theta
     \]

8. **Evaluate the integral**:
   - Separate the integrals:
     \[
     V = 2 \left( \int_{0}^{2\pi} d\theta \right) \left( \int_{0}^{1} (r - r^3) \, dr \right)
     \]
   - Compute the integrals:
     \[
     \int_{0}^{2\pi} d\theta = 2\pi
     \]
     \[
     \int_{0}^{1} (r - r^3) \, dr = \left[ \frac{r^2}{2} - \frac{r^4}{4} \right]_{0}^{1} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}
     \]
   - Combine the results:
     \[
     V = 2 \cdot 2\pi \cdot \frac{1}{4} = \pi
     \]

The final answer is \(\boxed{\pi}\)