1. **State the problem:** We are given a circle with center O and points A, B, C, D on the circumference. We know that $\angle BCD = 128^\circ$ and need to find $\angle OBD$.\n\n2. **Recall the properties:** \n- The angle at the center ($\angle BOD$) subtended by an arc is twice the angle at the circumference ($\angle BCD$) subtended by the same arc.\n- Radii of a circle are equal, so $OB = OD$, making triangle $OBD$ isosceles.\n\n3. **Calculate the central angle:**\n$$\angle BOD = 2 \times \angle BCD = 2 \times 128^\circ = 256^\circ$$\n\n4. **Adjust the central angle:**\nSince a full circle is $360^\circ$, the reflex angle $\angle BOD$ is actually the smaller angle around the circle, so we take\n$$\angle BOD = 360^\circ - 256^\circ = 104^\circ$$\n\n5. **Analyze triangle OBD:**\nTriangle $OBD$ has two equal sides $OB = OD$, so angles opposite these sides are equal. Let $\angle OBD = \angle ODB = x$.\n\n6. **Use triangle angle sum:**\nSum of angles in triangle $OBD$ is $180^\circ$:\n$$x + x + 104^\circ = 180^\circ$$\n$$2x = 180^\circ - 104^\circ = 76^\circ$$\n$$x = \frac{76^\circ}{2} = 38^\circ$$\n\n7. **Final answer:**\n$$\boxed{\angle OBD = 38^\circ}$$