For gravitational wave from twin stars, how was the tidal effect counted? As the primary indirect evidence, the work on calculating the rotational slow down earned the 1993 Nobel prize. However, I cannot find any where mention how the work deal with the tidal effect. Are both stars assumed to be perfect solid?
Any shape changes by gravitational effect will slow down the rotation and generate heat energy in the stars. Perfect solid means that the shape of star does not change at all by the gravitation of the other star, so there is no energy loss for rotation.
 A: First, let me explain the question a little further.  The Hulse-Taylor binary is a binary system composed of two neutron stars orbiting each other.  Each star is an extended body, and is in the gravitational field of the other, so should experience tidal forces, because one part of the star is closer than another to the opposite star, so the gravitational field is stronger.  [I'm using Newtonian terminology, but this made more rigorous using general relativity.]  Now, these tidal forces should raise bulges on each star, just as Earth's water bulges due to tidal forces from the Sun and Moon.  Of course, during the orbit these bulges must move around the stars, so there will be some loss of energy due to viscosity.  In addition, the bulges will typically be slightly offset from the line joining the two stars, so the gravity of these bulges themselves will also exert torques on the opposite stars.  Both these effects will sap energy from the orbital motion, which causes the orbit to shrink faster than would normally be the case if these tides did not exist.  [Note that this effect also exists for black holes, but will have a different size when matter is involved.]  In principle, this should leave an imprint on the data that we observe about the binary's orbit.
Now, to actually answer the question: Wikipedia tells me that the Hulse-Taylor binary has a maximum orbital velocity of just 450 km/s.  That's pretty fast, but it's still just 0.0015 times the speed of light (that's $v/c$).  This paper shows that the proportional effect of the tidal coupling on the phase is $\propto (v/c)^5 \approx 10^{-14}$.  Wikipedia also tells me that the orbital decay rate is measured as 0.997±0.002 times the predicted rate, which suggests that the tidal effects are tiny compared to the errors in that measurement.
So tidal effects theoretically contribute to the orbital decay of the Hulse-Taylor system, but it's at such a low level that we won't be able to measure it for a very long time.  As the orbit decays, the orbital velocity gets much larger, approaching 0.1 near merger, which means that the tidal effects will be proportionally more important, and thus potentially measurable.  But that's a very long way off.
