Question

The original question (now changed) read:

Let's consider

I will construct a closed simply connected manifold simply-connected $8$-manifold $M$ and a an $a\in H^k(M)$. Is it true H^3(M;\Bbb Z)$ such that the Poincare dual to $a$ is realisable as a immersed sphere or $a=bc$ for some $b,c\in H^∗(M)$?

I guess that $b,c\ne \pm 1$ is meant. Then $a\ne bc$ for any $a\in H^0(M)$. However, any nonzero non-generator in $H_2(S^2)$ is not realizable by a single immersed sphere. (A codimension zero immersion between manifolds must be a covering.) Also, any nonzero element in b$ of $H_4(\Bbb C P^2)$ a$ is not realizable by an immersed sphere.

OK, let's assume that $k>0$ is meant. Then the answer is yes for manifolds of dimension $\le 4$. Indeed, any $2$-dimensional homology class in a simply-connected closed map $4$-manifold is clearly representable by an embedded surfaceS^5\to M$, and hence (by cutting handles) by an immersed $2$-sphere.

But obviously the answer is no for manifolds of higher dimension. For instance, consider $a\in H^2(S^2\times S^2\times S^2)$ whose Poincare dual is represented by $S^2\times S^2\times pt$. If the latter homology class is representable by an immersed $4$-sphere, then the fundamental class of $S^2\times S^2$ can be represented by a map $f:S^4\to S^2\times S^2$. Then $f^*$ is trivial on $2$-cohomology and hence (by using the cup product) also on $4$-cohomology. But it shouldn't be, since Hom-dual element in $f$ is of degree one.

In response to edits 1,2: these look increasingly confusing H^5(M;\Bbb Z/2)$ to me. I can see the following meaningful question here: does there exist a closed simply-connected $m$-manifold $M$ and an $a\in H^k(M)$, \bmod 2$ reduction of $k\le m-1$, that on the one hand b$ is not a nontrivial product, and on the other hand there exists no immersion $f:S^k\to M$ such that $f^*(a)$ is a generator of $H^k(S^k)$? With integer coefficients, any $a$ of finite order would do, but with coefficients in a field (which as I understand are implicitly assumed in the edits 1,2) this makes sense. If this is not the question meant to be asked, I'm quitting and admit that my attempts at guessing failed.

Let $K$ be the suspension over $\Bbb C P^2$and let us consider $\Bbb Z/2$ coefficients. Then there is an $\alpha\in H^3(K)$ H^3(K;\Bbb Z/2)$ with $Sq^2(\alpha)\ne 0$, but the $\bmod 2$ cohomology ring of $K$ is trivial. Let $N$ be a regular neighborhood of a PL copy of $K$ in some $\Bbb R^m$. So $N$ is homotopy equivalent to $K$. A loop in $\partial N$ bounds a disk in $N$, which can be pushed off $K$ as long as $5+2\le m-1$. Thus $\partial N$ is simply-connected. Let $M$ be the double of $N$, i.e. $M=\partial (N\times I)$. So $M$ is a closed $m$-manifold, it is simply-connected by Seifert-van Kampen, and the inclusion $N\subset M$ is split by the projection $\phi:M\subset N\times I\to N$. So the cohomology of $N\simeq K$ is a direct summand in the cohomology of $M$. Let $\beta=\phi^\ast(\alpha)$, then $a:=Sq^2(\beta)\ne \gamma:=Sq^2\beta\ne 0$, and $a$ \gamma$ is not a nontrivial product. Since no the nonzero element of $H^5(S^5)$ H^5(S^5;\Bbb Z/2)$ is not in the image of $Sq^2$, there is no map $f:S^5\to M$ such that $f^*(a)\ne f^*(\gamma)\ne 0$.

Let $b\in H_5(M;\Bbb Z)$ be such that $\gamma(b)\ne 0$, and let $a\in H^{m-5}(M;\Bbb Z)$ be the Poincare dual of $b$. If $b$ is realized by an immersion, or just a map, $f:S^5\to M$, then $0\ne\gamma\smallfrown f_\ast[S^5]=f_\ast(f^\ast(\gamma)\smallfrown[S^5])$, contradicting $f^*(\gamma)=0$.

As for $m$, $\Bbb C P^2$ is the mapping cone of the Hopf map $h:S^3\to S^2$. The mapping cylinder of $h$ embeds in $S^2*S^3=S^6$, so $\Bbb CP^2$ embeds in $\Bbb R^7$ and $K$ in $\Bbb R^8$.