Does escape velocity depend on height and relationship between the period of rotation of earth and the period of revolution for a geostationary satellite?
Escape velocity ( in contrast to 'orbital velocity') is irrelevant in regards to geostationary satellite, by definition those are orbiting, and haven't escaped the Earth's gravity well. If by 'height' you mean the gravitational field potential 'F' at some distance 'r' ( F=G( m1*m2)/r^2 then yes, although the relative masses of Earth and some spacecraft at launch with the practical range of differences in 'r' on or near Earth's surface make the effect vanishingly small.
In some sense, the earth's rotation will give a benefit for any launch, because tangential velocity increases as one approaches the equator, but it doesn't affect the escape velocity of the Earth at all, since it entirely depends on the masses.
From https://en.wikipedia.org/wiki/Escape_velocity "In celestial mechanics, escape velocity or escape speed is the minimum speed needed for a free, non-propelled object to escape from the gravitational influence of a primary body, thus reaching an infinite distance from it. It is typically stated as an ideal speed, ignoring atmospheric friction. Although the term "escape velocity" is common, it is more accurately described as a speed than a velocity because it is independent of direction."
The escape velocity is independent of mass of the body and the direction of projection. It depends upon the gravitational potential at the point from where the body is launched. Since this potential depends on the height of the point of projection, the escape velocity depends on it. The higher the altitude the lesser the escape velocity required by the body to escape the earth's gravitational field. Geo-stationary satellites are those satellites which revolve along the equator of earth. This means that the time taken by them to complete a revolution is equal to the time taken by earth to complete one rotation which is equal to 1 day or 24 hours. The Earth rotates on its axis relative to the Sun every 24.0 hours mean solar time, with an inclination of 23.45 degrees from the plane of its orbit around the Sun. Mean solar time represents an average of the variations caused by Earth's non-circular orbit. T=2π√6.4×1069.8=5078s=84 minute 38s. It's just spinning at exactly the same rate as its orbit – one revolution every 27 days. Effectively, its day is as long as its year. This is no coincidence. It's called 'synchronous rotation' and is a result of the gravitational tug of war between the Earth and the Moon. Escape velocity decreases with altitude and is equal to the square root of 2 (or about 1.414) times the velocity necessary to maintain a circular orbit at the same altitude. At Earth's surface, if atmospheric resistance could be disregarded, escape velocity would be about 11.2 km (6.96 miles) per second. Since this potential depends slightly on the latitude and height of the point. Therefore, the escape velocity depends slightly on the height of the location from where the body is launched. Escape velocity decreases with altitude and is equal to the square root of 2 (or about 1.414) times the velocity necessary to maintain a circular orbit at the same altitude. At Earth's surface, if atmospheric resistance could be disregarded, escape velocity would be about 11.2 km (6.96 miles) per second.For the same reason that all objects, no matter their mass, fall at the same rate. In this case, the gravitational force isn't really relevant per se. The gravitational field strength is what is important. There is no altitude requirement in the definition for the escape velocity, but if the rocket hasn't risen above the sensible atmosphere there will be additional drag which, if not compensated for by continued thrust to sustain the achieved (escape) velocity, there will be a loss in velocity.