Let a is the first term and d is the common difference of the AP

So sum of p terms S_p = (p/2){2a + (p-1)d}

Sum of q terms S_q = (q/2){2a + (q-1)d}

Now since sum of p terms is q and sum of q terms is p,then

     S_p = S_q

=>  (p/2){2a + (p-1)d} = (q/2){2a + (q-1)d}

=> p{2a + (p-1)d} = q{2a + (q-1)d}

=> 2ap + p² d - pd = 2aq + q² d - qd

=> 2ap + p² d - pd - 2aq - q² d + qd = 0

=> 2a(p-q) + d(p² - d² ) - d(p-q) = 0

=>(p-q)*{2a + d(p+q) - d} = 0

=> (p-q)*{2a + d(p+q-1)} = 0

Now p-q = 0  ...1

and {2a + d(p+q-1)} = 0 ........2

Multiply (p+q)/2 in eqaution 2, we get

     {(p+q)/2}*{2a + d(p+q-1)} = 0

=> S_p+q = 0

So sum of p+q terms is 0.