The diagram subtracts the two-sample delayed branch from the direct branch: $y[n]=5x[n]-5x[n-2]$. Hence $H(e^{j\omega})=5(1-e^{-j2\omega})=10je^{-j\omega}\sin\omega$.
Its magnitude is $10|\sin\omega|$, zero at DC and at π, and maximal at π/2. It passes middle frequencies and rejects both extremes. Thus it approximates a **band-pass filter, C**.
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