I wrote following MATLAB code. Simulation results show the probability is around 0.602. I hope someone could confirm this with an analytic answer.
N = 1000000; A = rand(N, 1); B = rand(N, 1); C = rand(N, 1); p1 = 0.7; p2 = 0.6; c1 = rand(N, 1); c2 = rand(N, 1); ob1 = ((A > B) & (c1 < p1)) | ((A < B) & (c1 > p1)); ob2 = ((B > C) & (c2 < p2)) | ((B < C) & (c2 > p2)); ob = ob1 & ob2; pos = ob & (A > C); sum(pos) / sum(ob)
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I enumerate all the 6 possibilities of relative order of $A, B, C$. They all appear with probability 1/6.
The following lists shows with how much probability each case passes the two observers
$A>B>C$, $0.7\times 0.6$
$A>C>B$, $0.7\times 0.4$
$B>A>C$, $0.3\times 0.6$
$B>C>A$, $0.3\times 0.6$
$C>A>B$, $0.7\times 0.4$
$C>B>A$, $0.3\times 0.4$
Among them, $A>B>C$, $A>C>B$, $B>A>C$ are the valid cases. So
$\frac {0.7\times 0.6+0.7\times 0.4+0.3\times 0.6} {0.7\times 0.6+0.7\times 0.4+0.3\times 0.6+0.3\times 0.6+0.7\times 0.4+0.3\times 0.4} = 0.6027$