~ is equivalent to 'Not' in VB (VB uses Not for both logical and bitwise
negation).

As you have coded it, change the line
 B= B and [here I don't know what to do with ~] (1 << poss)
to
 B= B and (&hFF xor (1 << poss))

alternatively, the entire expression is

k=1<<poss
b=b or k
if value<>1 then b=b xor k

So this can be formulated as a function:

void SetBit(ref B, int poss, int Value) {
 B = (byte)(Value == 1 ? B | (1 << poss : B & ~(1 << poss);
}

I'd like Value to be a boolean, that way there is no room for the
"interpretation-error" in the above code if Value > 1 or < 0.

void SetBit(ref B, int poss, bool Value) {
 B = (byte)(Value ? B | (1 << poss : B & ~(1 << poss);
}

Rewrite the code to make it clear what the "?" operator is used for:

void SetBit(ref B, int poss, bool Value) {
 byte x;
 if ( Value )
   x = (byte)(B | (1 << poss));
 else
   x = (byte)(B & ~(1 << poss));
 B = x;
}

So, if the bit at poss is to be set, a 1 is shifted poss bits and or'ed
onto B. This clearly set's bit poss, and touches no other bit.

If the bit is to be cleared, B is and'ed with the bitwise invertion of 1
shifted poss bits (so it's a pattern of 1's, except in bit poss, where
it's a zero). This clearly removes any set bit poss, and leaves all
other bits intact.

Here is the way I think most clearly expresses what is (or should be :)
going on:

   static void SetBit(ref byte B, int poss, bool Value)
   {
     if ( Value )
       B |= (byte)(1 << poss); // or the bit onto B
     else
       B &= (byte)~(1 << poss); // mask the bit out of B
   }