This problem is asking for the major alkyne product that is formed when 1,1‑dichloro‑3‑methylbutane reacts with sodium amide (NaNH₂) ignoring any byproduct.
Thus, we can firstly say that such reaction of an alkyl halide with a strong base such as sodium amide is able to remove the halogen-based substituents and hence form insaturations in the form of double or triple bonds depending on the degree of substitution.
For instance, if one halogen substituent, then an alkene will be formed; but if two, then an alkyne (like this problem).
In such a way, as shown on the attached figure, the product will be 3-methylbutyne due to the loss of two chlorine substituents.
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