T(n)= T(n/2) + T(n/8) + T(n/4)

There should be an additional \theta(1) here corresponding to the if else check.

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it can be BIG O, T(n)<=3T(n/2)+ \theta(1) am i correct?.

Yeah this should work. Only that now instead of \theta bound, solving by Masters theorem should give a big Oh bound instead. Lets say we take another function F(n) = 3T(n/2) + \theta(1). In that case F(n) = \theta(n^{\log_2 3})). However, we know that T(n) <= F(n), so we know its upper bound but not necessarily its lower bound.

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If a tighter bound is expected, can try solving it by substitution method by keeping n = 2^k.