Electronic – Is self inductance and leakage inductance the same thing

Let's review the formulas that apply:

Weber-Turns (flux linkage): $$ \lambda=N\Phi $$

Flux: $$ \Phi=\frac {Ni} {\Re}$$

where \$\Re\$ is the reluctance, that depends on the coil's core material and geometry.

Voltage (emf)

$$ V=\frac {d\lambda}{dt}=\frac {N^2} {\Re} \frac {di}{dt}=L\frac {di}{dt}$$

The problem here is that it is not specified how the coil is excited.

Coil excited by a sinewave current

If it is excited by a sinewave current, then it is rather clear that lower inductance, implies lower flux linkage, lower emf. About the flux it is given by \$\Phi=\frac {Ni} {\Re}=\frac L N i\$. So reducing L alone doesn't give an indication of what \$\Phi\$ will do. For example, N can be reduced in more proportion than L if one choose a different coil material for example. Therefore \$\Phi\$ can be bigger or lesser than before.

Coil excited by a voltage

In contrast, if the coil excited by a voltage, then the flux linkage is given by the voltage alone (because \$V=\frac {d\lambda}{dt}\$, linkage is the voltage integration), and so doesn't depends on the inductance. Also the flux doesn't depends really on the inductance in this case, but just on the number of turns. (And because if wished you can have a big inductance with few turns by selecting a different core, then you might have a big or smaller flux by reducing L). Finally, again if excited by a voltage, the emf will be equal to it, so doesn't depends on the inductance.

Actually whatever excitation you consider, the first 3 options are wrong.

The last option asks about a steady current, so one can assume it is excited by a voltage in series with a resistor.

If self inductance is infinite, then any rate of change of flux would generate infinite voltage, but look at it the other way. Any amount of back emf can be generated with infinitesimally small rate of change of flux. When we apply a voltage to the primary of a transformer (with no secondary load), an equal and opposite back emf is initially generated and no current flows. In order to maintain the back emf, there must be a rate of change of flux, which requires an increasing primary current. The larger the self inductance, the smaller the rate of current rise required to maintain the emf. As self inductance approaches infinity the rate of rise of current approaches zero. So in the ideal transformer one could apply a voltage to the primary for ever with no current flow. The voltage across the socondary will be the primary voltage times the turns ratio. The current in the primary will be the secondary current / turns ratio.