Push and pop instruction of 8086 microprocessor?

I am not sure about 8086, but I can tell you the whole procedure in 8085.

PUSH instruction always pushes two bytes of data i.e. total 16 bits.

Example:

Assume that Stack is already initialized and SP is at 2008 address location. Then PUSH B instruction will have following steps:

1) The stack pointer (SP) will be pointing to the uppermost position of the stack (actually stack works in opposite order in terms of Addresses. e.g. if SP is now at address 2008, then PUSH instruction will store the contents on 2007 & 2006).

2) The contents of register B & C will be saved on to the stack such that contents of register B will be at 2007 & that of C will be at 2006 address location.

3) The SP is now modified to 2006.