Thinking process :
Based on the mirror isobar concept and binding energy concept.
Ans.
(a) According to question, a nuclide 1 is said to be mirror isobar of nuclide 2 , if $Z_1=N_2$ and $Z_2=N_1$.
Now in $ _{11} Na^{23}, Z_1=11, N_1=23-11=12$
$\therefore$ Mirror isobar of $ _{11} Na^{23}$ is $ _{12}, Mg^{23}$, for which $Z_2=12=N_1$ and $N_2=23-12=11=Z_1$
(b) As $ _{12}^{23} Mg$ contains even number of protons (12) against $ _{11}^{23} Na$ which has odd number of protons (11), therefore $ _{11}^{23} Mg$ has greater binding energy than $ _{11} Na^{23}$.