Step 1: Understand the question.
A cationic complex is one where the complex ion itself carries a positive charge. We check each compound to find one like that.
Step 2: Recall how to find the complex ion charge.
Split each compound into the counter ions outside the brackets and the complex inside. The charges must balance to zero overall.
Step 3: Test the first two.
In $Na_3[AlF_6]$, the three $Na^+$ leave the bracket with $[AlF_6]^{3-}$, which is negative. The compound $[Pt(NH_3)_2Cl_2]$ has no outside ions, so the complex is neutral.
Step 4: Test the third.
In $K_4[Fe(CN)_6]$, four $K^+$ leave $[Fe(CN)_6]^{4-}$, which is negative.
Step 5: Test the fourth.
In $[PtBr_2(NH_3)_4]Br_2$, two $Br^-$ sit outside. To balance them, the complex inside must be $[PtBr_2(NH_3)_4]^{2+}$, which is positive.
Step 6: Choose the answer.
The only cationic complex is option 4.
\[ \boxed{[PtBr_2(NH_3)_4]^{2+}} \]