A convex polyhedron $P$ has 26 vertices, 60 edges, and 36 faces, 24 of which are triangular, and 12 of which are quadrilaterals. A space diagonal is a line segment connecting two non-adjacent vertices that do not belong to the same face. How many space diagonals does $P$ have?
1. **Calculate the total number of line segments connecting any two vertices:**
   The total number of line segments connecting any two vertices in a polyhedron with \(26\) vertices is given by the binomial coefficient:
   \[
   \binom{26}{2} = \frac{26 \times 25}{2} = 325
   \]

2. **Subtract the number of edges:**
   We are given that the polyhedron has \(60\) edges. These edges are line segments connecting adjacent vertices, so we need to subtract these from the total number of line segments:
   \[
   325 - 60 = 265
   \]

3. **Subtract the diagonals of the quadrilateral faces:**
   Each quadrilateral face has \(4\) vertices and \(2\) diagonals. Since there are \(12\) quadrilateral faces, the total number of diagonals in these faces is:
   \[
   2 \times 12 = 24
   \]
   We need to subtract these \(24\) diagonals because they connect vertices that belong to the same face.

4. **Calculate the number of space diagonals:**
   Subtract the number of diagonals of the quadrilateral faces from the previous result:
   \[
   265 - 24 = 241
   \]

Thus, the number of space diagonals in the polyhedron is:
\[
\boxed{241}
\]