Concept:
Basicity depends on availability of lone pair
Step 1: In (a), N is sp$^3$ hybridized → lone pair freely available.
Step 2: In (b), lone pair is in aromatic ring → less available.
Step 3: In (c), lone pair is partially delocalized.
Step 4: In (d), O is more electronegative → holds lone pair tightly.
Conclusion:
Most basic = option (a)