The correct order for acid strength of compounds $$CH \equiv CH$$, $$CH_3 - C \equiv CH$$ and $$CH_2 = CH_2$$ is as follows:

First we recall that the acid strength of an organic molecule depends upon the stability of the conjugate base that is produced after loss of a proton. The more stable the conjugate base, the larger the acid dissociation constant $$K_a$$ and the smaller the $$pK_a\;(\;pK_a = -\log K_a\;),$$ hence the stronger the acid.

All the three compounds in the question lose a proton from a carbon atom, so we have to compare the stabilities of the corresponding carbanions:

$$CH \equiv CH \longrightarrow HC \equiv C^- + H^+$$

$$CH_3 - C \equiv CH \longrightarrow CH_3 - C \equiv C^- + H^+$$

$$CH_2 = CH_2 \longrightarrow CH_2 = CH^- + H^+$$

Now we analyse the factors that stabilise or destabilise each carbanion.

1. Hybridisation (s-character)
The electronegativity of a carbon atom increases with the percentage of s-character in its hybrid orbital. The order of s-character is

$$sp\;(50\%\,s) > sp^2\;(33\%\,s) > sp^3\;(25\%\,s).$$

Therefore a negative charge on an $$sp$$ carbon is held more tightly (more stabilised) than on an $$sp^2$$ carbon. Both alkynes have the negative charge on an $$sp$$ carbon, while the alkene places the charge on an $$sp^2$$ carbon. So immediately we get

$$\text{any terminal alkyne} \;\; > \;\; CH_2 = CH_2$$

in acid strength.

2. Inductive ( +I ) effect of the $$CH_3$$ group
Between the two terminal alkynes, $$CH_3 - C \equiv C^-$$ carries an additional $$CH_3$$ group next to the charged carbon. A methyl group exerts a +I (electron-releasing) effect which pushes electron density towards the negatively charged carbon, destabilising it.

In contrast, the carbanion $$HC \equiv C^-$$ has no such electron-releasing group; hence it is intrinsically more stable. Therefore

$$HC \equiv C^- \; >\; CH_3 - C \equiv C^-$$

in terms of stability, and consequently

$$HC \equiv CH \; >\; CH_3 - C \equiv CH$$

in acid strength.

3. Consolidating both factors
Combining the hybridisation argument (which places both alkynes ahead of the alkene) with the inductive argument (which distinguishes between the two alkynes), we obtain the complete order:

$$HC \equiv CH \; > \; CH_3 - C \equiv CH \; > \; CH_2 = CH_2.$$

Numerically this order is supported by the experimental $$pK_a$$ values:

$$pK_a(HC \equiv CH)\;\approx\;25,\qquad pK_a(CH_3 - C \equiv CH)\;\approx\;26,\qquad pK_a(CH_2 = CH_2)\;\approx\;44,$$

and the smaller the $$pK_a,$$ the stronger the acid.

Thus the correct sequence exactly matches Option B in the list.

Hence, the correct answer is Option B.