Endian-ness only comes into play when we break a 16-bit value into two 8-bit pieces, and there is a choice of approaches, which happens when such a 16-bit value is stored in memory so, for example:
With the encoding of immediates in instructions, such as ld bc, 0x1234: looking at the individual bytes of the machine code for this instruction, we will see 0x01 0x34 0x12, storing the 0x1234 in little endian byte order.
With a 16-bit value as data in memory, and looking at the individual bytes in memory, needs an endian-ness, so use little endian.
However, if the BC register pair is loaded with a value 0x1234, then that's the memory address that indirect use of BC will work with. There's no issue of endian-ness with this, since we're not looking at the individual bytes in this scenario.
Now, if you want to know how the BC register composes from B and C then we have to know which is the high order and which is the low order.
From the manual:
PUSH BC is PUSH B then C
So, that means that the stack memory will have the value from C then the value from B at the address +1 higher than where C was stored. Since 16-bit data in memory is stored using little endian byte order, we can conclude that C is the low order and B is the high order of the BC 16-bit register pair.
Thus, BC is B * 256 + C. Let's note though, that this is simply by definition, but doesn't really go to endian-ness — since there's no "addressing" of the individual bytes, there's also no other ordering as there is with memory (due to memory's having of both addresses and values).
However, the choice of which is high and which is low doesn't change that together they hold the value 0x1234, as per our examples.
