Hi,
Yes, one instruction cycle is 4 Clock cycles. That's where the Fosc/4 you see all around comes from. At 32MHz each instruction cycle (1/32M)*4 or 125ns.

Reading a whole port is obviously faster than Reading the 8 pins indvidually if that's what you're asking.

Four "sets" of those two line compiles to
Code:
000004                    M L00001
000004 8082               M         bsf     PORTC,  000h
000006 0E00               M         movlw   0
000008 B282               M         btfsc   PORTC, 001h
00000A 0E01               M         movlw   1
00000C 6E1A               M         movwf   _Sw1
00000E 8482               M         bsf     PORTC,  002h
000010 0E00               M         movlw   0
000012 B682               M         btfsc   PORTC, 003h
000014 0E01               M         movlw   1
000016 6E1B               M         movwf   _Sw2
000018 8882               M         bsf     PORTC,  004h
00001A 0E00               M         movlw   0
00001C BA82               M         btfsc   PORTC, 005h
00001E 0E01               M         movlw   1
000020 6E1C               M         movwf   _Sw3
000022 8082               M         bsf     PORTC,  000h
000024 0E00               M         movlw   0
000026 B282               M         btfsc   PORTC, 001h
000028 0E01               M         movlw   1
00002A 6E1D               M         movwf   _Sw4
00002C D7EB               M         bra     L00001
I count 20 instructions. Some of them (the btfsc) may take two cycles to execute but when they do the instruction AFTER the btfsc is skipped so in the end it'll still take 20 cycles, something I didn't take into account in the previous post. 20 * 125ns = 2.5us.

With that said I'm far from an expert in assembly programming so I truly hope someone that is corrects me if I'm wrong.

If you can explain what you're trying to do it may be easier to help you choose the best/fastest option.

/Henrik.