about the boot:
it loads at 0:7C00h by default. So it starts with org 7c00h.
Which place should I put my kernel? org 0 (0000:0000)?

org 230
mov ax,$ ;ax = $ = 230

org 100h
mov ax,$ ; ax= $ = 256

org 0
mov ax,$ ; ax=$ = 0

    

$: 230 
mov ax,$ ;ax = $ = 230 

$: 100h 
mov ax,$ ; ax= $ = 256 

$: 0 
mov ax,$ ; ax=$ = 0
    

The source operand specifies a 6-byte memory location that contains the base address (a linear address) and the limit (size of table in bytes) of the global descriptor table (GDT)…
…If the operand-size attribute is 16 bits, a 16-bit limit (lower 2 bytes) and a 24-bit base address (third, fourth, and fifth byte) are loaded. Here, the high-order byte of the operand is not used and the high-order byte of the base address in the GDTR or IDTR is filled with zeros.

3. The worse step:

Code:

gdtr: dw 23,gdtr-3
      db 0
      dw 0FFFFh,0,9A00h,0CFh
      dw 0FFFFh,0,9200h,0CFh    

GEeeeeezzzzzz what they meannnnnnn

but where in the code (bytes) it says to organize in some address?

I may be misunderstanding your question though.

you canuse several org inside the same source, because it will only change the base adress value.

when you say org xxxx, you just say to compiler a virtual base adress, something added to all label references.

org 100h
    nop
one:
    jmp two
    nop

    db 'one',0

;org 200h

    nop
two:
    jmp one
    nop

    db 'two',0    

org directive sets address at which the following code is expected to appear in memory. It should be followed by numerical expression specifying the address. This directive begins the new addressing space, the following code itself is not moved in any way, but all the labels defined within it and the value of $ symbol are affected as if it was put at the given address. However it's the responsibility of programmer to put the code at correct address at run-time.

org 100h        ; $     file
    nop         ; 100   00
one:
    jmp two     ; 101   01; two==0x201, jmp near
    nop         ; 104   04

    db 'one',0  ; 105   05

org 200h

    nop         ; 200   09
two:
    jmp one     ; 201   0A
    nop         ; 204   0D

    db 'two',0  ; 205   0E
                ; 209   12    

Think of first jump as jmp 201h.

So if you want to place code below org 200h at offset 200h, you must copy it yourself.

mov si,old_code_location
mov di,new_code_location
mov cx,length_of_code
cld
repnz movsb    

    db 'one',0  ; 105   05

; label for source address should be placed here
; because it should refer to address where the code is located now
old_code_location:
org 200h
; label for destination address should be placed here
; because it should be at 200h
new_code_location:
    nop         ; 200   09
two:
    jmp one     ; 201   0A
    nop         ; 204   0D

    db 'two',0  ; 205   0E
                ; 209   12
length_of_code = $-new_code_location    

i don't understand why you don't understand that....

org 100h
nop
db 'aaa',0
org 200h
nop
db 'bbb',0    

0D346:0100 : 90 61 61 61 00 90 62 62 62 00 .aaa..bbb.    

0D346:0100 : 90 61 61 61 00 .aaa.
(...)
0D346:0200 : 90 62 62 62 00 .bbb.    

What i dont understand is why in debug it showns datas at same code place, if we org it in another place.

org directive sets address at which the following code is expected to appear in memory. It should be followed by numerical expression specifying the address. This directive begins the new addressing space, the following code itself is not moved in any way, but all the labels defined within it and the value of $ symbol are affected as if it was put at the given address. However it's the responsibility of programmer to put the code at correct address at run–time.

but all the labels defined within it and the value of $ symbol are affected as if it was put at the given address

org 0x100

mov si,str1
mov di,str2

str1 db 'test1'
str2 db 'test2'