Which has higher rate of hydrogenation - methyl propene or trans-2-butene?
We should be able to determine which molecule will hydrogenate faster by comparing the activation energies for hydrogenation of these two molecules.  A catalyst does not alter the reaction mechanism, it merely reduces the activation energy for the process.

The heat of formation of 2-methyl-propene is ca. 1.3 kcal/mol higher than that of trans-2-butene, presumably because of the adverse steric interaction between the geminal methyl groups in 2-methyl-propene.  The transition state for this reaction will start to resemble the corresponding alkane in its eclipsed conformation (due to the cis-hydrogen delivery).  If we only consider methyl-methyl steric interactions, then for 2-methyl-propene as we approach the transition state the two geminal methyl groups, initially separated by a 120 degree ($\ce{sp^2}$) angle, will displace towards a 109 degree ($\ce{sp^3}$) angular separation.  Therefore, the methyl-methyl steric interaction should increase in 2-methyl-propene as we approach the transition state.  2-methyl-propene's  transition state will be more destabilized than the ground state, consequently increasing the activation energy required to reach it.  With trans-2-butene, there are no methyl-methyl steric interactions in the ground state or transition state; its activation energy, the energy separating the ground state and transition state, should not be increased or decreased.

Based on this analysis, the ground state - transition state energy separation would be expected to increase (slightly) for 2-methyl-propene as compared to trans-2-butene.  Consequently, faster hydrogenation of trans-2-butene would be predicted.  However the effect is expected to be very small.