Full-bridge output doesn`t work as expected
You need to remove the previous loads and redesign your filter. The initial values of 39.8u and 1.6u are for a filter with Q=1 (note: underdamped), considering a load of 5 Ω (V(i)). Combining them into a bridge results in the inductors and capacitors being in series, thus the equivalent values will double for the inductors and halven for the capacitors. The output load will have to be doubled if you want the same response (V(a2,b2)).
If you leave the previous loads there while adding the 8 Ω load, the equivalent load will be the two 5 Ω loads in series, all in parallel with the 8 Ω, or about 4.44 Ω. Combine this with 2*39.8u and 1.6u/2 and you get a filter with a Q=0.446, which is well overdamped (V(a,b)).
If you remove the 5 Ω loads and leave the 8 ω, without modifying the LC values, you get a filter with a Q=0.8 (V(x,y)), and only if you change the load to be 10 Ω you will have the original Q=1 (V(a2,b2)).
Finally, if your intention is to have the load 8 Ω and the filter with Q=1, then you have to recalculate the LC values to be L=31.8u and C=2u (V(p,q)).