Determining the reduction potential of a half cell reaction at a given pH
The K you refer to relates to the overall cell reaction and should be Q, the reaction quotient. This is only equal to K when the potential difference between the two 1/2 cells has fallen to zero.

This question does not refer to a full cell, but to a 1/2 cell. The form of The Nernst Equation you must use, assuming we are at 298K, is:

$\rm E=E^\circ+\dfrac{0.0591}{z}log\dfrac{[oxidised\,\, form]}{[reduced\,\, form]}$

Where z is the number of moles of electrons transferred which, in this case = 2.

This becomes:

$\rm E=E^\circ+\dfrac{0.0591}{2}log[Cu^{2+}]$

The 1/2 cell in question is:

$\rm Cu_{(aq)}^{2+}+2e\rightleftharpoons Cu_{(s)}\,\,\,\,\,\,\,\,\,\,E^\circ=+0.34\,V$

At such a high pH the copper(II) ions will precipitate.

$\rm pH=14\,\,\,so\,\,\,pOH=0$

So $\rm c[OH^-]=1\,\,mol/l$

This will precipitate out the copper(II) ions as the hydroxide.

So we have:

$\rm Cu(OH)_{2(s)}\rightleftharpoons Cu_{(aq)}^{2+}+2OH_{(aq)}^-$

$\rm K_{sp}=[Cu^{2+}][OH^-]^2=10^{-19}$

$[\ce{OH-}]$ is 1 (pH is 14), so $[\ce{Cu^2+}]$ is equal to $\pu{e-19}$.

As you can see this is a very low concentration. Applying Le Chatelier's Principle to the 1/2 cell couple we would predict that reducing the concentration of Cu(II) will drive the position of equilibrium to the left. This would push out electrons and make the electrode potential of the 1/2 cell more negative.

Putting in the numbers into the Nernst Equation:

$\rm E = 0.34+\dfrac{0.0591}{2}\log \pu{e-19}$

$\rm E = 0.34- 1.12\,\,\,\,\,V$

$\rm E = -0.78\,\,\,\,V$

As predicted, it has been reduced considerably.