Step 1: Concept
Convert the complex number into polar form.
Step 2: Meaning
Observe that
\[
1+\sin\alpha+i\cos\alpha
=
(1+\sin\alpha)+i\cos\alpha.
\]
Step 3: Analysis
Its modulus is
\[
r=\sqrt{(1+\sin\alpha)^2+\cos^2\alpha}
=\sqrt{2(1+\sin\alpha)}
=2\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right).
\]
Its argument is
\[
\phi=\frac{\pi}{4}+\frac{\alpha}{2}.
\]
Hence
\[
1+\sin\alpha+i\cos\alpha
=
2\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)
\left(\cos\phi+i\sin\phi\right).
\]
Applying De Moivre's theorem,
\[
A=
2^8\cos^8\left(\frac{\pi}{4}-\frac{\alpha}{2}\right),
\]
and the argument becomes
\[
8\phi
=
2\pi-4\alpha.
\]
Therefore the principal argument is
\[
-4\alpha.
\]
Step 4: Conclusion
Thus
\[
A=
2^8\cos^8\left(\frac{\pi}{4}-\frac{\alpha}{2}\right),
\qquad
\theta=-4\alpha.
\]
Final Answer: (B)