This appears to be the entry to a routine, allocating some space on
the stack. It's rather bad code because of the lack of optimization,
but...
The first two instructions appear to set the space requirements on the
stack (15 bytes).
The two shifts are to mask off the bottom four bits, IOW, to truncate
it to a multiple of 16 bytes (to maintain stack alignment). The right
shift shifts the four bottom bits off the right, the left shift shifts
four zero bits back in.
The truncation from the two shifts, and the third instruction combine
to round the space requirements *up*, so that the required number of
16-byte chunks (one, in this case) to hold the required locals (15
bytes) is computed.
The last instruction actually allocates the space on the stack.
At even -O1, I'd expect most of that to be collapse to one or two
instructions, perhaps as little as a "sub esp,16", or even omitted
entirely if the optimization can eliminate the need for a stack
frame..