Step-by-step Solution:

1. Understand the structure of allene ($C_3H_4$). Allene is an organic compound with the chemical formula $C_3H_4$. It is the simplest member of a class of hydrocarbons called allenes, which feature two adjacent carbon-carbon double bonds. The structure of allene can be drawn as: $H_2\overset{1}{C}=\overset{2}{C}=\overset{3}{C}H_2$ We need to determine the hybridization of each carbon atom: the two terminal carbons ($C_1$ and $C_3$) and the central carbon ($C_2$).

2. Determine hybridization based on bonding. The hybridization of an atom in a molecule can be determined by counting the number of sigma ($\sigma$) bonds and lone pairs of electrons around it. This sum gives the number of hybrid orbitals required.

   • Sum of 4 $\rightarrow$ $sp^3$ hybridization (e.g., methane)
   • Sum of 3 $\rightarrow$ $sp^2$ hybridization (e.g., ethene)
   • Sum of 2 $\rightarrow$ $sp$ hybridization (e.g., ethyne)

   In hydrocarbons, carbon atoms do not have lone pairs, so we only need to count the number of sigma bonds.

3. Analyze the terminal carbon atoms ($C_1$ and $C_3$). Let's consider the leftmost carbon atom, $C_1$. It is bonded to:

   • Two hydrogen atoms via single bonds (2 $\sigma$ bonds).
   • The central carbon atom ($C_2$) via a double bond. A double bond consists of one $\sigma$ bond and one pi ($\pi$) bond.

   So, the total number of $\sigma$ bonds around $C_1$ is $2$ (from C-H bonds) + $1$ (from C=C bond) = 3 $\sigma$ bonds. Since there are 3 sigma bonds and 0 lone pairs, the total number of electron domains is 3. This corresponds to $sp^2$ hybridization for $C_1$. By symmetry, the rightmost carbon atom, $C_3$, is also bonded to two hydrogen atoms and the central carbon, so it also has $sp^2$ hybridization.

4. Analyze the central carbon atom ($C_2$). The central carbon atom, $C_2$, is bonded to:

   • The left carbon atom ($C_1$) via a double bond (1 $\sigma$ bond, 1 $\pi$ bond).
   • The right carbon atom ($C_3$) via a double bond (1 $\sigma$ bond, 1 $\pi$ bond).

   So, the total number of $\sigma$ bonds around $C_2$ is $1$ (from $C_1=C_2$) + $1$ (from $C_2=C_3$) = 2 $\sigma$ bonds. Since there are 2 sigma bonds and 0 lone pairs, the total number of electron domains is 2. This corresponds to $sp$ hybridization for $C_2$.

5. Conclusion. The carbon atoms in allene ($C_3H_4$) exhibit two types of hybridization:

   • Terminal carbons ($C_1, C_3$): $sp^2$
   • Central carbon ($C_2$): $sp$

   Therefore, the types of hybridization present are $sp$ and $sp^2$.

6. Match with options. Comparing our findings with the given options:

   • A: sp and sp3 - Incorrect.
   • B: sp and sp2 - Correct.
   • C: only sp3 - Incorrect.
   • D: sp2 and sp3 - Incorrect.

The correct option is B.