Spectroscopy of excited state lithium
The energy levels are in increasing energy  3S, 3P, 3D and the final state is 2P. The P-P transition is not allowed by $\Delta L =\pm1$ selection rule. The two transitions are 3S-2P at approx 812 nm and 3D-2P at approx 610 nm. As you are asked about the lowest energy transition this must be the 3S-2P. (Li also has non-zero nuclear spin depending  on its isotope which lead to hyperfine transitions and are ignored ) 

With some spin-orbit coupling the (Russell-Saunders) terms symbols are for the upper level $^2S_{1/2}, $ and for the lower level $^2P_{3/2}, \, ^2P_{1/2}$. The $M_J$ levels are assumed to be split so the degeneracy is 2 for the S and 2 or 4 for the P states as given by $2J+1$.  The selection rules on $M_J$ levels are $M_J = 0,\, \pm 1$.

Thus there are four transitions possible from $^2S_{1/2}$ to $^2P_{1/2}$ starting in  $  M_J = 1/2 $ in $^2S$ to $(1/2, -1/2)$ in $^2P$ and similarly $  M_J = -1/2 \rightarrow (1/2,\, -1/2)$, making 4 transitions in all as the splitting of the S and P levels will be different. (This is the same case as the splitting of sodium $D_1$ line but 'upside down' in energy to that case). 

The other transitions are from $^2S_{1/2}$ to $^2P_{3/2}$. In this case there are 6 transitions as the P splits into 4 levels. The transition between the $\pm 1/2$ make four of these (as just outlined above) and in addition there are transitions between $M_J = 3/2 \rightarrow 1/2$ and $M_J = -3/2 \rightarrow -1/2$.

This makes ten transitions in all, thus it seems I have calculated the wrong transitions (or misunderstood completely),  however, you should be able to follow the method. Note that the observed spectral lines may not correspond to transition as distinct transitions may have the same energy gaps.

I've added a sketch below, it is hand drawn but should be clear enough. The $S_{1/2} - P_{1/2}$  transitions are a subset of these as you can see.