You are confusing the speed of the bike (mph), with the speed of the engine.(rpm)
I said that if a bike is at the HP peak, then for that road speed, the bike is putting the most torque to the back wheel that it can.
An example:
I'm feeling Buellish, so we'll use an XB12R.
Data for an XB12R:
89.8Hp. @ 6,700rpm
72.6LbFt. @ 5,750rpm
Rear tire: 180/55ZR-17
This isn't entirely exact, but the conversion from rear wheel rpm to mph for such a tire is roughly mph = 0.0738 * rwrpm
Pick any road speed you want. I'm going to go with 40 mph.
In order to go 40 mph, the rear tire must be turning at 40/0.0738 = ~542 rpm
Assuming full throttle...
If you are going 40 mph, and are at the torque peak, the torque being applied to the rear wheel is
72.6 * ( 5750 / 542 ) = 770 lbft
At the horsepower peak, the Buell is putting out
( 89.8 * 5252 ) / 6700 = ~ 70.4 lbft from the engine
but the torque to the rear wheel for 40 mph at the HP peak is
70.4 * ( 6700 /542 ) = 870 lbft
Repeat the calcs for other speeds/bikes/whatever. It doesn't matter.
For a given road speed, you will put the most torque to the back wheel if you are at the HP peak.
-- Michael