Yes it is: Rotating the table by the angles $\frac{360^{\circ}}{11}, 2 \cdot \frac{360^{\circ}}{11}, 3 \cdot \frac{360^{\circ}}{11}, \ldots, 10 \cdot \frac{360^{\circ}}{11}$, we obtain 10 new positions of the table. By the assumption, it is obvious that every one of the 11 members of the Committee will be seated in front of the card with his name in exactly one of these 10 positions. Then by the Pigeonhole Principle there should exist one among these 10 positions in which at least two of the $11(>10)$ members of the Committee will be placed in their positions, as claimed.