1. Use the fact that $P(1,1)$ is the circumcentre

Since $P$ is the circumcentre of triangle $ABC$, it is equidistant from $A,B,C$. So, $PA=PB=PC.$

Given: $A(a,3),\quad B(b,5),\quad C(a,b),\quad P(1,1).$

2. Equate $PA^2$ and $PC^2$

$PA^2=(a-1)^2+(3-1)^2=(a-1)^2+4$ $PC^2=(a-1)^2+(b-1)^2$

Thus, $(a-1)^2+4=(a-1)^2+(b-1)^2$ $(b-1)^2=4$ $b-1=\pm 2$ $b=3 \text{ or } -1.$

But $ab>0$, and we will also use the next condition.

3. Equate $PB^2$ and $PC^2$

$PB^2=(b-1)^2+(5-1)^2=(b-1)^2+16$ $PC^2=(a-1)^2+(b-1)^2$

Hence, $(b-1)^2+16=(a-1)^2+(b-1)^2$ $(a-1)^2=16$ $a-1=\pm 4$ $a=5 \text{ or } -3.$

Since $ab>0$, valid positive pairing is $a=5,\quad b=3.$ (The pair $a=-3, b=-1$ also gives $ab>0$, but let us check the geometry: the final result turns out same in this setup? We proceed with the valid option-producing pair from the circumcentre condition and test.)

So the triangle vertices are $A(5,3),\quad B(3,5),\quad C(5,3).$

This makes $A$ and $C$ coincide, which is impossible for a triangle. So we must carefully use all pairwise equalities with the original coordinates.

4. Correct use: equate $PA^2$ and $PB^2$

$PA^2=(a-1)^2+4$ $PB^2=(b-1)^2+16$

Thus, $(a-1)^2+4=(b-1)^2+16$ $(a-1)^2-(b-1)^2=12.$

Also from $PA^2=PC^2$: $(a-1)^2+4=(a-1)^2+(b-1)^2$ $(b-1)^2=4.$

So, $b=3 \text{ or } -1.$

Substitute into $(a-1)^2-(b-1)^2=12:$ $(a-1)^2-4=12$ $(a-1)^2=16$ $a=5 \text{ or } -3.$

Now possible pairs with $ab>0$ are: $(a,b)=(5,3) \quad \text{or} \quad (-3,-1).$

But $(5,3)$ makes $A=(5,3)$ and $C=(5,3)$, degenerate triangle. Hence reject it.

Therefore, $a=-3,\quad b=-1.$

Then $A(-3,3),\quad B(-1,5),\quad C(-3,-1).$

Check: $PA^2=(-4)^2+2^2=20,$ $PB^2=(-2)^2+4^2=20,$ $PC^2=(-4)^2+(-2)^2=20,$ so this is correct.

5. Find equation of line $AP$

Points $A(-3,3)$ and $P(1,1)$.

Slope: $m_{AP}=\frac{1-3}{1-(-3)}=\frac{-2}{4}=-\frac12.$

Equation: $y-1=-\frac12(x-1).$

Multiply by 2: $2y-2=-x+1$ $x+2y-3=0.$

So line $AP$ is $x+2y-3=0.$

6. Find equation of line $BC$

Points $B(-1,5)$ and $C(-3,-1)$.

Slope: $m_{BC}=\frac{-1-5}{-3-(-1)}=\frac{-6}{-2}=3.$

Equation through $B(-1,5)$: $y-5=3(x+1)$ $y=3x+8.$

So line $BC$ is $y=3x+8.$

7. Find intersection $Q(k_1,k_2)$ of $AP$ and $BC$

Solve $x+2y-3=0$ and $y=3x+8.$

Substitute: $x+2(3x+8)-3=0$ $x+6x+16-3=0$ $7x+13=0$ $x=-\frac{13}{7}.$

Then $y=3\left(-\frac{13}{7}\right)+8=-\frac{39}{7}+\frac{56}{7}=\frac{17}{7}.$

Thus, $Q\left(-\frac{13}{7},\frac{17}{7}\right).$

So, $k_1+k_2=-\frac{13}{7}+\frac{17}{7}=\frac{4}{7}.$

8. Final answer

$k_1+k_2=\frac{4}{7}.$ So the correct option is B.