It looks like in your third step you tried to take the square root of both sides (since $\sqrt{12}\approx 3.4641$). However, multiple things went wrong.
First, on the left hand side it is not true that $\sqrt{x^2+y^2}=x+y$. For example, take $x=3,y=4$; then $\sqrt{x^2+y^2}=5$ but $x+y=7$. You can't move the "$+$" through the "$\sqrt{\cdot}$." In your case, we can't take the square root of $a^2+1$ and get $a+1$.
Second, on the right hand side you can't just ignore the $a$. We have $$\sqrt{12a}=\sqrt{12}\times\sqrt{a},$$ not $$\sqrt{12a}=\sqrt{12}\times a.$$ This is a variant of the same mistake above: you can't move the "$\times$" out of the "$\sqrt{\cdot}$."
Square-rooting both sides of the equation $a^2+1=12a$ isn't going to help: if done correctly we get $$\sqrt{a^2+1}=\sqrt{12}\times\sqrt{a},$$ which is true but ... just awful. But there's something else you can do here which will (HINT: think about moving the terms around a bit).
If $p=q=0$, then any $(\theta,\phi)$ is a solution.
If $p\not=0$ and $q=0$, then $\theta=\frac{\pi}{2}+k\pi$ with any $\phi$ where $k$ is an integer.
If $q\not=0$, then there is no solution.
Proof :
If $p\not=0$ and $q=0$, then we have $\sin^4\theta=1$ and $\cos^4\theta=0$, so $\theta=\frac{\pi}{2}+k\pi$ with any $\phi$ where $k$ is an integer.
In the following, $q\not=0$.
$$p\sin^{4}{\theta}-q\sin^{4}{\phi}=p\tag1$$
$$p\cos^{4}{\theta}-q\cos^{4}{\phi}=q\tag2$$
From $(1)$, we have
$$p(1-\sin^4\theta)=-q\sin^4\phi\tag3$$
From $(2)$, we have
$$p\cos^4\theta=q(1+\cos^4\phi)\tag4$$
$(3)\times \cos^4\theta-(4)\times (1-\sin^4\theta)$ gives
$$0=-q\sin^4\phi\cos^4\theta-q(1+\cos^4\phi)(1-\sin^4\theta)\tag5$$
Dividing the both sides of $(5)$ by $q\not=0$, we have
$$-\sin^4\phi\cos^4\theta=(1+\cos^4\phi)(1-\sin^4\theta)\tag6$$
Suppose that $\sin\phi=0$. Then, since $p\cos^4\theta=2q$, we have $p\not=0$, so dividing the both sides of $p\sin^4\theta=p$ by $p$, we get $\sin\theta=\pm 1$ which implies $2q=p\cos^4\theta=0$, a contradiction.
Suppose that $\cos\theta=0$. Then, we have $\cos^4\phi=-1$ which is impossible.
So, we have $\sin\phi\cos\theta\not=0$.
Then, dividing the both sides of $(6)$ by $\sin^4\phi\cos^4\theta\not=0$, we have
$$-1=\frac{1+\cos^4\phi}{\sin^4\phi}\cdot \frac{1-\sin^4\theta}{\cos^4\theta}$$
which is impossible since LHS is negative while RHS is non-negative.$\ \blacksquare$
Best Answer
It is much simpler to deal with this as follows:\begin{align}\frac{\sin(x)}{\cos(x)}=\cos(x)&\iff\sin(x)=\cos^2(x)\\&\iff\sin(x)=1-\sin^2(x)\\&\iff\sin^2(x)+\sin(x)-1=0.\end{align}So, solve the equation $y^2+y-1=0$.