Observe the given structure of 1-buten-3-yne where specific bonds...

The given molecule is 1-buten-3-yne: $\text{CH}_2=\text{CH}-\text{C}\equiv\text{C}-\text{H}$.

First, determine the hybridization of each carbon atom:
The terminal carbon $C_1$ ($\text{CH}_2=$) is bonded to three atoms with no lone pairs, so it is $sp^2$ hybridized.
The carbon $C_2$ ($=\text{CH}-$) is bonded to three atoms, so it is $sp^2$ hybridized.
The carbon $C_3$ ($-\text{C}\equiv$) is bonded to two atoms, so it is $sp$ hybridized.
The carbon $C_4$ ($\equiv\text{C}-\text{H}$) is bonded to two atoms, so it is $sp$ hybridized.

Now, analyze the labeled bonds:
Bond 'a' is a double bond between $C_1$ and $C_2$. A double bond consists of one $\sigma$ bond and one $\pi$ bond. (A $\rightarrow$ III)
Bond 'b' is a single bond between $C_2$ ($sp^2$)$and C_3$ ($sp$). It is a $\sigma$ bond formed by the overlap of an $sp^2$ orbital and an $sp$ orbital. (B $\rightarrow$ II)
Bond 'c' is a triple bond between $C_3$ and $C_4$. A triple bond consists of one $\sigma$ bond and two $\pi$ bonds. (C $\rightarrow$ I)
Bond 'd' is a single bond between $C_4$ ($sp$) and a hydrogen atom ($s$). It is a $\sigma$ bond formed by the overlap of an $sp$ orbital and an $s$ orbital. (D $\rightarrow$ IV)

Therefore, the correct matching is A-III, B-II, C-I, D-IV.

Answer: A-III, B-II, C-I, D-IV