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You wish to make substitutions in all except Subscript expressions I believe. This can be done by leveraging the precedence of replacement rules, like this:

{p, Subscript[p, 1], Subscript[p, 2]} /. {x_Subscript :> x, p -> 1}

{1, Subscript[p, 1], Subscript[p, 2]}

Due to the traversal order of ReplaceAll the rule x_Subscript :> x acts first, "replacing" any expression with the head Subscript by itself. Since expressions that have already been replaced are not matched again, the second rule p -> 1 is never applied to these subexpressions.

You wish to make substitutions in all except Subscript expressions I believe. This can be done by leveraging the precedence of replacement rules, like this:

{p, Subscript[p, 1], Subscript[p, 2]} /. {x_Subscript :> x, p -> 1}

{1, Subscript[p, 1], Subscript[p, 2]}

Due to the traversal order of ReplaceAll the rule x_Subscript :> x acts first, "replacing" any expression with the head Subscript by itself. Since expressions that have already been replaced are not matched again, the second rule p -> 1 is never applied to these subexpressions.

You wish to make substitutions in all except Subscript expressions I believe. This can be done by leveraging the precedence of replacement rules, like this:

{p, Subscript[p, 1], Subscript[p, 2]} /. {x_Subscript :> x, p -> 1}

{1, Subscript[p, 1], Subscript[p, 2]}

The rule x_Subscript :> x acts first, "replacing" any expression with the head Subscript by itself. Since expressions that have already been replaced are not matched again, the second rule p -> 1 is never applied to these subexpressions.

You wish to make substitutions in all except Subscript expressions I believe. This can be done by leveraging the precedence of replacement rules, like this:

{p, Subscript[p, 1], Subscript[p, 2]} /. {x_Subscript :> x, p -> 1}

{1, Subscript[p, 1], Subscript[p, 2]}

Due to the traversal order of ReplaceAll the rule x_Subscript :> x acts first, "replacing" any expression with the head Subscript by itself. Since expressions that have already been replaced are not matched again, the second rule p -> 1 is never applied to these subexpressions.