This is a mov instruction, not a ld instruction.
There is no loading going on. The value in EBX is already loaded. Otherwise it wouldn't be in a register.
It's hard for me to take someone seriously who doesn't understand the difference between a load and a move but feels as if they have the right to an opinion about assembly.
https://cdrdv2-public.intel.com/819723/325462-sdm-vol-1-2abc... Volume 1, page 7-3 (page 181 of the PDF).
In the context of assembly, load/store refers to memory ⇆ register.
https://www.felixcloutier.com/x86/mov-2: "MOV — Move to/from Debug Registers ... At the opcode level, the reg field within the ModR/M byte specifies which of the debug registers is loaded or read. The two bits in the mod field are ignored. The r/m field specifies the general-purpose register loaded or read."
The question is whether "Intel also uses 'load' in the descriptions of register-to-register moves."
What Felix Cloutier says is not probative on this question.
What Intel says in Intel's manual posted on Intel's website is the definitive authority on what Intel says on the matter.
https://cdrdv2-public.intel.com/819723/325462-sdm-vol-1-2abc..., Vol. 2B page 4-42, (physical page 1263): "MOV—Move to/from Debug Registers ... At the opcode level, the reg field within the ModR/M byte specifies which of the debug registers is loaded or read. The two bits in the mod field are ignored. The r/m field specifies the general-purpose register loaded or read."