Selecting from Point column in PostgreSQL?

You seem to be using the native PostgreSQL POINT type rather than the PostGIS GEOMETRY(POINT); this answer is for the PostGIS extension! See the answer in the cross post on DBA.SE for the PostgreSQL native type.

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You will need to to construct an actual geometry from your input coordinates to compare against values in a GEOMETRY column, e.g.

• ST_SetSRID(ST_MakePoint(93.2321, 21.0321)) or

• 'POINT(93.2321 21.0321)'::GEOMETRY(POINT, <SRID)

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However:

PostGIS stores geometry with standard IEEE floating point precision, and checking 4 decimal places against 15 leaves plenty of room for misalignment.

Thus, you'd also want to make sure you are comparing precision in the same order of magnitude; use e.g. ST_Snap or ST_SnapToGrid to make sure two points are considered equal even if they differ after the 4th decimal place

Finally, the recommended way to compare PostGIS geometries is ST_Equals.

SELECT *
FROM   indsolv.address
WHERE  ST_Equals(
         a.location,
         ST_Snap(
           a.location,
           ST_SetSRID(ST_MakePoint(93.2321, 21.0321), <srid>),
           0.0001
         )
       )
;

Refer:
SELECT a.* FROM indsolv.address a WHERE a.location ~= point '(93.2321, 21.0321)';

... WHERE a.location ~= point '93.2321, 21.0321';
... WHERE a.location ~= '93.2321, 21.0321'::point;
... WHERE a.location ~= '(93.2321, 21.0321)'::point;
... WHERE a.location ~= cast('93.2321, 21.0321' AS point);
... WHERE a.location ~= cast('(93.2321, 21.0321)' AS point);

... WHERE a.location ~= '(93.2321, 21.0321)';
... WHERE a.location ~= '93.2321, 21.0321';

... WHERE a.location ~= point('93.2321, 21.0321)';
... WHERE a.location ~= point('(93.2321, 21.0321)');
... WHERE a.location ~= point('93.2321', '21.0321');