1. We compare the first ionization enthalpies of the 3d-series elements: Ti, Mn, Ni, and Zn.

2. General trend across a period:

   • First ionization enthalpy generally increases from left to right due to increasing effective nuclear charge.
   • But in transition metals, the trend is not perfectly regular because of shielding and extra stability of certain electronic configurations.

3. Electronic configurations:

   • $\mathrm{Ti}: [\mathrm{Ar}]\,3d^2 4s^2$
   • $\mathrm{Mn}: [\mathrm{Ar}]\,3d^5 4s^2$
   • $\mathrm{Ni}: [\mathrm{Ar}]\,3d^8 4s^2$
   • $\mathrm{Zn}: [\mathrm{Ar}]\,3d^{10}4s^2$

4. Relative comparison:

   • $\mathrm{Ti}$ is toward the left, so it has the lowest first ionization enthalpy among these.
   • $\mathrm{Mn}$ has a higher value than Ti because of greater effective nuclear charge and the stability associated with half-filled $3d^5$.
   • $\mathrm{Ni}$ lies further right, so its first ionization enthalpy is higher than Mn.
   • $\mathrm{Zn}$ has a completely filled $3d^{10}$ subshell and relatively high effective nuclear charge, giving it the highest first ionization enthalpy among these.

5. Thus, the increasing order is $\mathrm{Ti} < \mathrm{Mn} < \mathrm{Ni} < \mathrm{Zn}$

6. Match with options:

   • A: $\mathrm{Ti} < \mathrm{Mn} < \mathrm{Zn} < \mathrm{Ni}$ ❌
   • B: $\mathrm{Zn} < \mathrm{Ni} < \mathrm{Mn} < \mathrm{Ti}$ ❌
   • C: $\mathrm{Mn} < \mathrm{Ti} < \mathrm{Zn} < \mathrm{Ni}$ ❌
   • D: $\mathrm{Ti} < \mathrm{Mn} < \mathrm{Ni} < \mathrm{Zn}$ ✅

Therefore, the correct option is D.