Members KJ and KL by inspection are zero force. there may be more but not immediately evident to mho.

I would say that you can divide them into two groups:

At nodes with two member and no external force

• Node C: CD,CB,
• Node G: FG, GH
• Node K: KJ, KL

If you write down the equilibrium at node C for x and y you would get:

$$x: quad sum F_x = 0 rightarrow CD = 0 $$ $$y: quad sum F_y = 0 rightarrow CB = 0 $$

Same at the other nodes. Note that nodes A, and I are not in this category because of the reaction forces,

nodes with 3 members and no external force

• Node M: bar DM
• Node O: bar FO
• Node P: bar NP

In those nodes what happens, is that if you take the y-axis, since you don't have an external force you get:

$$y: quad sum F_y = 0 rightarrow DM = 0 $$

Basically, in those situations, the node DM offers stability when the other BM and MN are in compression (and will demonstrate a nominal force), while in the case of tension there will be no force on DM.

Notice than EN is not same. If EN is removed then the distribution of the forces will change entirely. The point of the exercise is to think which bar will have zero (or close to zero) loads.