In pipeline ideally $CPI=1$
So in $1$ cycle $1$ instruction gets completed
Throughput is instructions in unit time
In pipeline $1,$ cycle time$=$ max stage delay $=800\ \text{psec}$
In $800\ \text{psec},$ we expect to finish $1$ instruction
So, in $1\;\text{ps},$ $\dfrac{1}{800}$ instructions are expected to be completed, which is also the throughput for pipeline $1.$
Similarly pipeline $2,$ throughput$=\dfrac{1}{600}$
Throughput increase in percentage
$=\dfrac{\text{new-old} }{\text{old}}\times100$
$= \dfrac{\dfrac{1}{600}-\dfrac{1}{800}}{\dfrac{1}{800} }\times 100$
$=\dfrac{200}{600}\times 100$
$=33.33 \%$