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Given two equality constraints: x+y==250 and z+p==65 where x=190, y=50, z=45, p=15, I want to specify an error-tolerance level of e=0.05 below which the first equality should be TRUE and above which the second equality be FALSE.

x=190, y=50, z=45, p=15;
eq1= x+y==250; (* e=0.0416 is the percentage change from 250*)
eq2= z+p==65; (* e=0.0833 is the percentage change from 65*)

(*Mathematica output*)
(*FALSE, FALSE*)

(*I like to receive for given e=0.05*)
(*TRUE, FALSE*)

How can I set the error-tolerance level of e=0.05 for the two constraints?

Any idea?

ClearAll[choppedEqual]
SetAttributes[choppedEqual, HoldFirst, Listable]
choppedEqual[a_ == b_, c_] := Chop[a - N @ b, b c] == 0.;

Examples:

choppedEqual[x + y == 250, .0416]

True

choppedEqual[x + p == 65, .0833]

False

choppedEqual[x + y == 250, x + p == 65, 0.0416, 0.0833]

True, False

Some more ways, with the relative error e = 0.05:

Block[Internal`$EqualTolerance = MachinePrecision + Log10[e],
 x + y == 250., z + p == 65. (* advantage: equations written in terms of == *)
 ]
(* True, False *)

SetPrecision[x + y, -Log10[e]] == SetPrecision[250, -Log10[e]], 
 SetPrecision[z + p, -Log10[e]] == SetPrecision[65, -Log10[e]]
(* True, False *)

svn = NDSolve`ScaledVectorNorm[Infinity, e, 0];
svn[x + y - 250, 250] < 1, svn[z + p - 65, 65] < 1
(* True, False *)

You can use Congruent.

ClearAll[Congruent]
TOL = 0.05;
Congruent[a_, b_] := 
 If[Abs[a - b] > 0, Abs[a - b] / Norm[a, b, Infinity] <= TOL, True];

Addendum

your answer does not show how I test these equations with Congruent. Can you show me how I use Congruent?

Please note that lines 37 and 39 on the attached screenshot use Congruent.

It can be seen in the linked function page that the infix symbol of Congruent ("≡") can be entered as "Esc === Esc".

x + y ≡ 250 
(* True *)

z + p ≡ 65 
(* False *)

Using Block we can also make bulk evaluations of many equalities.

Block[Equal = Congruent,
 x + y == 250, z + p == 65
]

(* True, False *)