Answer: The possible values are \(\boxed{k=\dfrac52}\) or \(\boxed{k=3}\).

Let \(v_1\) be the speed of \(P\) immediately before the collision at the lowest point.

Initially the string is horizontal, and at the lowest point the particle has fallen a vertical distance \(a\). By conservation of energy,

\[\frac12mv_1^2=\frac12m(12ag)+mga.\]

So

\[v_1^2=12ag+2ag=14ag.\]

During the collision, \(P\) and \(Q\) coalesce. The combined mass is

\[(1+k)m.\]

Let their common speed just after the collision be \(v_2\). Conservation of momentum gives

\[(1+k)mv_2=mv_1.\]

Therefore

\[v_2=\frac{v_1}{1+k}.\]

Now find the tension before and after the collision. At the lowest point, the radial equation before the collision is

\[T_1-mg=\frac{mv_1^2}{a}.\]

Using \(v_1^2=14ag\),

\[T_1=mg+\frac{m(14ag)}{a}=15mg.\]

After the collision, the radial equation is

\[T_2-(1+k)mg=\frac{(1+k)mv_2^2}{a}.\]

Hence

\[T_2=(1+k)mg+\frac{(1+k)m}{a}\left(\frac{v_1}{1+k}\right)^2.\]

Using \(v_1^2=14ag\),

\[T_2=(1+k)mg+\frac{14mg}{1+k}.\]

The tension immediately after the collision is half the tension immediately before the collision, so

\[(1+k)mg+\frac{14mg}{1+k}=\frac12(15mg).\]

Cancel \(mg\):

\[(1+k)+\frac{14}{1+k}=\frac{15}{2}.\]

Multiply by \(1+k\):

\[(1+k)^2+14=\frac{15}{2}(1+k).\]

Multiply by \(2\):

\[2(1+k)^2+28=15(1+k).\]

Expand and simplify:

\[2k^2-11k+15=0.\]

Factorise:

\[(2k-5)(k-3)=0.\]

Therefore

\[\boxed{k=\frac52\quad\text{or}\quad k=3}.\]