The bmp file has no padding bytes, and its size matches the image dimensions (4374054 = 810*1800*3 +0x36 for the header).
On the other hand, applying an LSB filter reveals that something is wrong on the left side of the image (the 630 first columns from the left look filled with random bits).
The three colors are affected in the same way, and the second LSB is normal. So, we certainly have an LSB encoding with one bit per byte.

The fact that the bits form a rectangle suggests the encoding was done following the image order rather than the file order. The two most logical choices (for occidentals) are left-to-right and up-to-down. I was going for the former; the grace of the random Bug made me do the latter first.

Here is how the data begin:

00 02 eb 9b 78 9c d4 b9 65 54 ...

The index of coincidence reveals a flat distribution. The data could be either encrypted or compressed. But then, the two first bytes are suspiciously low.
0x2eb9b (191387) is also very close to the rectangle size: 630*810*3 / 8 = 191362. And as it happens, 78 9C is a typical beginning for strings compressed with zlib (deflate algorithm).

Quote from http://garethrees.org/2007/11/14/pngcrush/ :

The header byte 78 meaning “deflate compression with a 32 KiB window”.
The informational byte 9c meaning “the default compression algorithm was used” (plus a checksum).”

So all that has to be done is to extract the least significant bits in column-major (up-to-down), skip the first four bytes indicating the size of the file, and decompress the rest with zlib.

The output file is a pdf describing a few products from SCIOS. This file is the flag.

import sys, zlib,Image, struct
bmp = Image.open("sp113.bmp")
pix = bmp.load()
lsb = []
for x in range(640):
    for y in range(810):
        lsb.extend( str(i&1) for i in pix[(x,y)] )
lsb = "".join(lsb)
lsb = "".join(chr(int(lsb[i:i+8],2)) for i in range(0,len(lsb),8))
length = struct.unpack(">I",lsb[:4])[0]
pdf = zlib.decompress(lsb[4:4+length])
outfile = open("sp113.pdf","wb") #"b" is for windows users