If there are M stages and N instructions then without pipelining time taken will be = CPI$_{\text{Non-pipeline}}$*T$_{\text{Non-pipeline}}$
there are M stages ==> if you take T$_{\text{Non-pipeline}}$ = M*T$_{\text{pipeline}}$, then CPI$_{\text{Non-pipeline}}$=1
Total Time in Non-Pipeline Model = M*T$_{\text{pipeline}}$ = M clock cycles where these clocks are of PIPELINE model
Given that,in pipelining where 33% instructions are branch instructions, which will cause 3 stall cycles
∴ 67% are Normal instructions which will take CPI = 1 and 33% instructions will take CPI = (3+1) = 4
Total Cycles in Pipeline Model = (% normal ) * (time for normal instructions) + (% for branch ) * (time for Branch instructions)
= 0.67 * 1 + 0.33 * 4 = 0.67 + 1.32 = 1.99
so speed up = $\frac{\text{Cycles in non-pipeline model}}{\text{Cycles in non-pipeline model}}$ = $\frac{M}{1.99}$ = 8
so M = 8*1.99 which is nearly 16 .