How to Redraw or Interpret This
To avoid confusions, you can mark all the nodes in the circuit first. We know currents from different branches either enter or leave from a node. Also keep in mind the KCL. The total current entering a node = the total current leaving a node. Let us mark these in the circuit.

You can arbitarily assume and mark the direction of all branch currents at each node. Because in the end when we calculations the values of currents , we will end up with its original directions (-ve sign means, the direction is actually opposite to what we assumed).
Now before you proceed, you can formulate some current relations in the 
circuit using KCL.
As you can see:

At node A : \$ I_1 = I_2 + I_4 \$

At node B : \$ I_4 = I_5 + I_6 \$

At node C : \$ I_3 = I_5 + I_6 \$

At node D : \$ I_7 = I_2 + I_3 \$

From these we can deduce that \$ I_4 = I_3 , I_7 = I_1 = I_2 + I_3 \$. Using 
KVL we can find all branch currents with these relations. 
If you want to use 'loop current' method to solve this circuit. You gotta assume 3 loop currents, for three closed loops you see there. Again solving for R5,R6,R7,R8 into a single equivalent resistance \$Z = ((R5 + R7)||R6) + R8 \$ , will reduce the circuit to a simpler one with only two loop currents to solve.