$m = -5 = 1011$
$-m = 5 = 0101$
$r \ = -3 = 1101$
let x and y represent the number of bits in m and r.
$A = 1011 \ 0000 \ 0$
$S = 0101 \ 0000 \ 0$
$P = 0000 \ 1101 \ 0$
Least significant 2 bits of P are $10$
1) if $10$ then $P + S$, $0101 \ 1101 \ 0$ and shift 1 bit to the right, So P = $0010 \ 1110 \ 1$
2) if $01$ then $P + A$, $1101 \ 1110 \ 1$ and shift 1 bit to the right, So P = $1110 \ 1111 \ 0$
3) if $10$ then $P + S$, $0011 \ 1111 \ 0$ and shift 1 bit to the right, So P = $0001 \ 1111 \ 1$
4) if $11$ then do nothing and shift 1 bit to the right, So P = $ 0000 \ 1111 \ 1$
After dropping the LSB, we're left with $0000 \ 1111$ which is $15$