Number of address bits = 16.
#Tag bits = 9.
#Offset bits = 3 (block/line size is 8 bytes and byte-addressing assumed).
So, number of set bits = 16 - 9 - 3 = 4.
Number of sets $=2^4=16$.
No. of cache blocks/lines $=\frac{\text{Cache size}}{\text{Block size}} \\=\frac{1024}{8} = 128.$
So, we have 128 cache blocks and 16 sets meaning $\frac{128}{16} = 8$ way associative cache.