Ana had Php A to begin with
Bill had Php B to begin with
Claire had Php C to begin with

When Ana gave Bill B, then Bill had 2B
When Ana gave Claire C, then Claire had 2C
And Ana was left with A-B-C

When Bill gave Ana A-B-C, then Ana had 2A-2B-2C
When Bill gave Claire 2C, then Claire had 4C
And Bill was left with 2B-(A-B-C)-2C

Simplifying what Bill was left with:
2B-(A-B-C)-2C
2B-A+B+C-2C
-A+3B-C

So Bill was left with -A+3B-C

When Claire gave Ana 2A-2B-2C, then Ana had 4A-4B-4C
When Claire gave Bill -A+3B-C, then Bill had 
-2A+6B-2C
And Claire was left with 4C-(2A-2B-2C)-(-A+3B-C)

Simplifying what Claire was left with:

4C-(2A-2B-2C)-(-A+3B-C)
4C-2A+2B+2C+A-3B+C
-A-B+7C

Since each had Php 52

Ana had Php 52, so

     4A-4B-4C = 52
 

----------------------------

Bill had Php 52, so

    -2A+6B-2C = 52

-----------------------------

Claire had 52, so 

     -A-B+7C = 52


So we have this

     4A-4B-4C = 52
    -2A+6B-2C = 52
     -A- B+7C = 52

Can you solve that system for A, B, and C.
If not, post again asking how to solve it.

Answers A = 84.5, B = 45.5,  C = 26

Checking:

Ana had Php 84.5
Bill had Php 45.5
Claire had Php 26

When Ana gave Bill Php 45.5, then Bill had Php 91
When Ana gave Claire Php 26, then Claire had Php 52
And Ana was left with Php 84.5-45.5-26 or Php 13 

When Bill gave Ana Php 13, then Ana had Php 26
When Bill gave Claire Php 52, then Claire had Php 104
And Bill was left with Php 91-13-52 or Php 26

When Claire gave Ana Php 26, then Ana had Php 52
When Claire gave Bill Php 26, then Bill had Php 52
And Claire was left with Php 104-26-26 or Php 52

So it is correct

Edwin