CPU TIME $(T) =$ No. of Instructions $( I ) \times$ No. of Cycles Per Instruction $(c) \times$ Cycle Time $(t)$
OR
CPU TIME $(T) = \dfrac{\text{No. of Instructions(I) $\times$ No. of Cycles Per Instruction (c)}}{\text{Clock frequency (f)}}$
$\rightarrow T = I_{c} \times CPI \times F^{-1}$
$\rightarrow \dfrac{T \times F}{CPI} = I_{c}$
$P_1$ & $P_2$ executing same instruction set So,
No. of Instructions same for both $= I_1 = I_2 = I$
If $P_1$ takes $T_1$ time,
$\rightarrow T_2 = 0.75\times T_1 \rightarrow\dfrac{T_{2}}{ T_{1}}=0.75$
If $P_1$ incurs $C_1$ clock cycles per instruction,
$\rightarrow C_2 =1.2 \times C_1\rightarrow \dfrac{C_{2}}{C_{1}}=1.2$
Since $I$ is same for both,
$\rightarrow \dfrac{ ( f_{1} \times T_{1} )}{c1} = \dfrac{ ( f_{2} \times T_{2} )}{c2}$ and $f_1 =1\ GHz$
$\rightarrow F_2 =(\dfrac{C_{2}}{C_{1}}) \times (\dfrac{T_{1}}{T_{2}}) \times F_{1}$
$= \dfrac{1.2 \times 1 GHz}{0.75}=1.6\ GHz$
Hence, the clock frequency of $P_2$ is $=1.6\ GHz$.