Step 1: Use the binomial theorem.
Since $I$ and $A$ commute, \[ (2I+A)^4 = \sum_{k=0}^{4}\binom{4}{k}(2I)^{4-k}A^k \]
Step 2: Replace each power of A.
From $A^2=A$ we get $A^k=A$ for $k \geq 1$. The $k=0$ term is a pure $I$ term.
Step 3: Add the terms.
The $k=0$ term is $2^4 I = 16I$. The terms with $k \geq 1$ all multiply $A$: \[ \left[\binom{4}{1}2^3+\binom{4}{2}2^2+\binom{4}{3}2+\binom{4}{4}\right]A = (32+24+8+1)A = 65A \]
Step 4: Finish.
So $(2I+A)^4 = 16I+65A$. Subtracting $65A$ leaves $16I$. This is option 2.
Step 5: Check with special matrices.
Idempotent matrices include the identity $I$ and the zero matrix $O$. For $A=I$: $(3I)^4 - 65I = 81I-65I = 16I$. For $A=O$: $(2I)^4 - O = 16I$. Both give $16I$, which agrees with our answer. Option 1 ($8I$) fails both tests. Option 3 gives $17I$ for $A=I$ and $16I$ for $A=O$, so it is not a fixed value. Option 4 gives $9I$ for $A=I$, so it fails.
Final Answer:
The answer is $16I$.
\[ \boxed{16I} \]