Please don't take this as gospel, but I think that your bias supply, since it's negative voltage, sees the elevated ground, thereby making it *more* negative. Since you're using a 50volt zener, the math is about right.
I am still working my way through Merlin's Power Supply book, so I could be totally wrong. 
Look at it this way:
1. Due to the number of turns in the winding of the PT, the voltage relationship between the CT and either end of the PT secondary is fixed.
2. Normally, you connect the CT to ground (0v) as a reference for the secondary. To lower the effective B+, you can place the zener diode in between the CT and ground. This moves the reference voltage of the CT.
3. Typically, the zener is oriented so that the CT is connected to an end which places the CT at an effective
negative voltage; that is, for a 50v zener, the CT thinks it is attached to -50v instead of 0v. The secondary still outputs all the voltage it always did, but since the CT is referenced to a negative voltage, the rectifier output is lowered by the amount of the shifted reference.
4. The bias, if derived from a bias tap, now has the zener voltage added to the original voltage. The bias supply is probably still referenced to 0v; if it were referenced to the zener, it might not develop enough bias.
5. Even if you reference the bias to the zener, you still have shifting voltage problems. Until the amp is taken off standby, there might not be enough current flow through the zener to define its zener voltage, and therefore drop the B+ (or add to the bias). So with this arrangement, even if you meticulously reference to the zener, you might still have bias shift.
Which is why you do it Sluckey's way. Or don't use zeners to drop B+, but that's a different can of worms and maybe more $$.