The period of the same pendulum will be longer on Mt. Everest than somewhere closer to the ground.
Let
[tex]T \approx 2\; \pi \sqrt{\dfrac{L}{g}}[/tex] if the pendulum swings at small angles. In other words, reducing the value of [tex]g[/tex] increases the length of the period.
How does the value of [tex]g[/tex] compare on Mt. Everest and at sea level?
[tex]g = \dfrac{G \cdot M}{r^{2}}[/tex],
where
[tex]r[/tex] is at the denominator. A large value of [tex]r[/tex] will lead to a small value of [tex]g[/tex]. Mt. Everest is further away from the center of the earth than a spot at sea level. As a result, [tex]g[/tex] will be larger at the sea level and smaller on top of Mt. Everest.
Now, back to the approximation
[tex]T \approx 2\; \pi \sqrt{\dfrac{L}{g}}[/tex].
The value of [tex]g[/tex] is smaller on Mt. Everest than at sea level. As a result, the time period of the pendulum [tex]T[/tex] will be larger on Mt. Everest than at sea level.
