The equivalent resistance between A and B is

A. $36\Omega $ if ${{V}_{A}}\rangle {{V}_{B}}$
B. $18\Omega $ if ${{V}_{A}}\rangle {{V}_{B}}$
C. 0 if ${{V}_{A}}\rangle {{V}_{B}}$
D. $54\Omega $ if ${{V}_{A}}\rangle {{V}_{B}}$

Hint: We know that when diodes are present in a circuit, biasing has a key role. So we have two cases: one where A is positive and B negative and the second where A is negative and B is positive. The options are given for the first case, so we could bias accordingly and hence find the equivalent resistance.
Formula used:
Equivalent resistance in series,
${{R}_{eq}}={{R}_{1}}+{{R}_{2}}$

Complete answer:
In the question, we are given a circuit consisting of three diodes and two resistances and we are supposed to find the equivalent resistance across the terminals A and B.
Since the circuit consists of diodes we have two conditions: ${{V}_{A}}\rangle {{V}_{B}}$ and ${{V}_{A}}\langle {{V}_{B}}$
For ${{V}_{A}}\rangle {{V}_{B}}$,

We see that when${{V}_{A}}\rangle {{V}_{B}}$ , all the three diodes are forward biased and will act as a short circuit. So, under this condition the circuit could be redrawn as,

Since the resistors provide opposition to the current flow, the current would take the easiest path without any resistance if available, that is, APQB. So, the equivalent resistance when ${{V}_{A}}\rangle {{V}_{B}}$ would be zero.

Hence, option C is the correct answer.

Additional information:
When ${{V}_{A}}\langle {{V}_{B}}$
Now the biasing would be,

Here, we see that diodes ${{D}_{1}}$ and ${{D}_{3}}$ are reverse biased and would act like open circuits and hence there wouldn’t be any current flow in those arms. But the diode ${{D}_{2}}$ is forward biased and will act as a short circuit. Circuit could now be redrawn as,

Now across the terminals A and B we have two resistors connected in series and the equivalent resistance will now be,
$R=18\Omega +36\Omega =54\Omega $

Note:
Basically, we should be careful about the biasing when diodes are given in the circuit as they may act as short and open circuits accordingly. If you make a simple mistake even in assigning the signs you would end up getting the whole thing wrong. Also, always remember that the current will always take the easiest path.