Step 1: Use a different Hermitian test: real expectation values.
An operator $\hat{X}$ is Hermitian exactly when $\langle\psi|\hat{X}|\psi\rangle$ comes out real for every possible state $|\psi\rangle$, not just for one special state. This gives a route that does not need adjoint bookkeeping term by term.
Step 2: Write down the expectation value of $\hat{A}$.
Since $\hat{A}$ is not Hermitian, its expectation value in a general state need not be real; call it $\langle\psi|\hat{A}|\psi\rangle = z$, a general complex number, $z = x+iy$ with $x,y$ real. A standard identity for any operator gives $\langle\psi|\hat{A}^{\dagger}|\psi\rangle = \langle\psi|\hat{A}|\psi\rangle^{*} = z^{*} = x - iy$.
Step 3: Expectation value of the combination.
For $\hat{X} = c\hat{A}-d\hat{A}^{\dagger}$:
\[ \langle\psi|\hat{X}|\psi\rangle = cz - dz^{*} = c(x+iy) - d(x-iy) = (c-d)x + i(c+d)y \]
Step 4: Demand this stays real for every $x,y$.
Since $x$ and $y$ range over all real numbers as the state $|\psi\rangle$ varies, the coefficient of $x$ and the coefficient of $iy$ must each be real on their own:
\[ (c-d) \text{ must be a real number}, \qquad (c+d) \text{ must be purely imaginary} \]
Step 5: Test each option against both conditions.
Option (A), $c=i,d=i$: $c-d=0$ (real), $c+d=2i$ (purely imaginary). Both conditions hold.
Option (B), $c=1,d=1$: $c-d=0$ (real), but $c+d=2$ is a plain real number, not purely imaginary. Fails.
Option (C), $c=-1,d=i$: $c-d=-1-i$, which is not real at all. Fails already.
Option (D), $c=i,d=-i$: $c-d=2i$, not real. Fails.
Final Answer:
Only $c=i$, $d=i$ keeps every expectation value of $c\hat{A}-d\hat{A}^{\dagger}$ real, so this is the Hermitian choice.
\[ \boxed{c=i,\ d=i} \]