P0 = string1[z] & 0x3F | 0x80;

Whast do You think, how will this be interpreted if the OR is done first and right after that the AND. I am too old to remember the order of execution of those and, or, xor and so on - so I use prantheses - always when unclear.

It is my assumption that in this case the OR is actually performed first and then the AND cancels the effect of it. This turns the bit 7 down which in turn resets Your display.

The () parentheses do not cost CPU-wise anything but they are a sort of guarantee of execution order - which is really important in this case.

P0 = (string1[z] & 0x3F) | 0x80; // Ahhhh.... lot better