The NFA represented by P, accepts string “00” and then at final state (other than initial state) we have self loop of “1” , so we conclude that it must accept the string of the form of $\rightarrow$ ϵ + 0 X* 01*, where X is regular expression (01*1 + 00 ) {resolving the loop at middle state}. It matches with statement 1.
Similarly, The NFA represented by Q, has the form of$\rightarrow$ ϵ + 0X*0, where X is regular expression (10*1 + 00 ) {resolving the loop at middle state}. It matches with statement 2.
The NFA represented by R, has the form of $\rightarrow$ ϵ + 0X*1, where X is regular expression (10*1 + 01 ) {resolving the loop at middle state}. It matches with statement 3.
The NFA represented by S, accepts string “01” and then at final state (other than initial state) we have self loop of “0” , so we conclude that it must accept the string of the form of $\rightarrow$ ϵ + 0X* 10*, where X is regular expression (10*1 + 10 ) {resolving the loop at middle state}. It matches with statement 4.