The matrix $A = \begin{pmatrix} 3 & -b \\ -2 & -3 \end{pmatrix}$ has only one eigenvalue when the discriminant of its characteristic equation is zero.

The characteristic equation is given by:

$\text{det}(A - \lambda I) = 0$

For matrix $A$, the characteristic polynomial is:

$\text{det}\begin{pmatrix} 3 - \lambda & -b \\ -2 & -3 - \lambda \end{pmatrix} = 0$

Expanding the determinant:

$(3 - \lambda)(-3 - \lambda) - (-b)(-2) = 0$

Simplifying:

$(9 + 3\lambda - 3\lambda - \lambda^2) - 2b = 0$

This simplifies to:

$9 - \lambda^2 - 2b = 0$

Thus, the characteristic polynomial is:

$\lambda^2 - 9 + 2b = 0$

For the matrix to have only one eigenvalue, the discriminant of this quadratic equation must be zero. The discriminant of $\lambda^2 + 0\lambda + (-9 + 2b) = 0$ is:

$\Delta = 0^2 - 4(1)(-9 + 2b) = 0$

This gives:

$36 - 8b = 0$

Solving for $b$:

$8b = 36$

$b = \frac{36}{8} = 4.5$

Thus, the value of $b$ is $\boxed{4.5}$.