The low 8 bits of SI, DI, BP and SP weren't accessible before, but now they are in 64-bit mode.

The earliest ancestor of x86 was the CPU of the Datapoint 2200 terminal, implemented originally as a board of TTL logic chips and then by Intel in a single chip (the 8008). On that architecture, there was only a single addressing mode for memory: it used two 8-bit registers "H" and "L" to provide the high and low byte of the address to be accessed.

Next came the 8080, which provided some more convenient memory access instructions, but the HL register pair was still important for all the old instructions that took up most of the opcode space. And the 8086 was designed to be somewhat compatible with the 8080, allowing automatic translation of 8080 assembly code.

16-bit x86 didn't yet allow all GPRs to be used for addressing, only BX or BP as "base", and SI/DI as "index" (no scaling either). BP, SI and DI were 16-bit registers with no equivalent on the 8080, but BX took the place of the HL register pair, that's why it can be accessed as high and low byte.

Also the low 8 bits of the x86 flag register (Sign,Zero,always 0,AuxCarry,always 0,Parity,always 1,Carry) are exactly identical to those of the 8080 - that's why those reserved bits are there, and why the LAHF and SAHF instructions exist. The 8080 "PUSH PSW" (Z80 "PUSH AF") instruction pushed the A register and flags to the stack, so LAHF + PUSH AX emulates that (although the byte order is swapped, with flags in the high byte whereas it's the low byte on the 8080).

Fun fact, that obviously you already know but may be interesting to others.

In the encoding the registers are ordered AX, CX, DX, BX to match the order of the 8080 registers AF, BC (which the Z80 uses as count register for the DJNZ instruction, similar to x86 LOOP), DE and HL (which like BX could be used to address memory).